Combination of Resistors — NEET Physics MCQs with Solutions
Free NEET Physics Combination of Resistors MCQs with step-by-step solutions (59 questions). Part of Current Electricity. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Combination of Resistors · easy · theory
The alloys constantan and manganin are used to make standard resistance due to they have
A. Low temperature coefficient of resistance
B. Low resistivity
C. Both high resistivity and low temperature coefficient of resistance ✓ Correct
D. High resistivity
Solution: A standard resistance needs a large resistance in a small length (high resistivity) whose value hardly changes with temperature (low temperature coefficient); manganin and constantan have both.
Q2 — Combination of Resistors · easy · theory
When a potential difference is applied across the ends of a linear metallic conductor
A. The free electrons are accelerated continuously from the higher potential end to the lower potential end of the conductor
B. The free electrons are set in motion from their position of rest
C. The free electrons are accelerated continuously from the lower potential end to the higher potential end of the conductor
D. The free electrons acquire a constant drift velocity from the lower potential end to the higher potential end of the conductor ✓ Correct
Solution: Collisions with the lattice prevent continuous acceleration, so the electrons settle to a steady drift velocity, moving from the low-potential end towards the high-potential end.
Q3 — Combination of Resistors · easy · theory
The electric resistance of a certain wire of iron is $R$. If its length and radius are both doubled, then
A. The resistance and the specific resistance, will both remain unchanged
B. The resistance will be doubled and the specific resistance will be halved
C. The resistance will be halved and the specific resistance will remain unchanged ✓ Correct
D. The resistance will be halved and the specific resistance will be doubled
Solution: $R' = \rho(2l)/[\pi(2r)^{2}] = \rho(2l)/(4\pi r^{2}) = R/2$, and specific resistance depends only on the material, so it is unchanged.
Q4 — Combination of Resistors · easy · numerical
A thick wire is stretched so that its length become two times. Assuming that there is no change in its density, then what is the ratio of change in resistance of wire to the initial resistance of wire
A. $1 : 4$
B. $4 : 1$
C. $3 : 1$ ✓ Correct
D. $2 : 1$
Solution: With the density (and hence the volume) unchanged, $R \propto l^{2}$, so the new resistance is $4R$. The change is $4R - R = 3R$, giving a ratio $3 : 1$.
Q5 — Combination of Resistors · easy · numerical
The length of the resistance wire is increased by 10%. What is the corresponding change in the resistance of wire
A. 21% ✓ Correct
B. 9%
C. 10%
D. 25%
Solution: On stretching, $R \propto l^{2}$, so the new resistance is $(1.1)^{2} = 1.21$ times the old — an increase of 21%.
Q6 — Combination of Resistors · easy · numerical
Two wires that are made up of two different materials whose specific resistance are in the ratio $2 : 3$, length $3 : 4$ and area $4 : 5$. The ratio of their resistances is
A. $1 : 2$
B. $5 : 8$ ✓ Correct
C. $6 : 8$
D. $6 : 5$
Solution: $R = \rho l/A$, so $R_1 : R_2 = (2 \times 3/4) : (3 \times 4/5) = 1.5 : 2.4 = 5 : 8$.
Q7 — Combination of Resistors · easy · numerical
A wire has a resistance of 12 ohm. It is bent in the form of equilateral triangle. The effective resistance between any two corners of the triangle is
A. 12 ohms
B. 9 ohms
C. 6 ohms
D. 8/3 ohms ✓ Correct
Solution: Each side has resistance $12/3 = 4\ \Omega$. Between two corners one side (4 $\Omega$) is in parallel with the other two in series (8 $\Omega$): $4 \times 8/12 = 8/3\ \Omega$.
Q8 — Combination of Resistors · easy · numerical
Two resistances are joined in parallel whose resultant is $\dfrac{6}{8}$ ohm. One of the resistance wire is broken and the effective resistance becomes $2\ \Omega$. Then the resistance in ohm of the wire that got broken was
A. $3$
B. $\dfrac{3}{5}$
C. $2$
D. $\dfrac{6}{5}$ ✓ Correct
Solution: The unbroken wire is the one left carrying the current, so it is $2\ \Omega$. From $\dfrac{8}{6} = \dfrac{1}{2} + \dfrac{1}{x}$ we get $\dfrac{1}{x} = \dfrac{5}{6}$, i.e. $x = 6/5\ \Omega$.
Q9 — Combination of Resistors · easy · theory
Lamps used for household lighting are connected in
A. Parallel ✓ Correct
B. Series
C. Mixed circuit
D. None of the above
Solution: In parallel each lamp gets the full mains voltage and can be switched on or off independently of the others.
Q10 — Combination of Resistors · easy · theory
The equivalent resistance of resistors connected in series is always
A. Equal to sum of component resistors ✓ Correct
B. Equal to the mean of component resistors
C. Less than the lowest of component resistors
D. In between the lowest and the highest of component resistors
Solution: In series the same current passes through each resistor and the potential differences add, so $R_{eq} = R_1 + R_2 + \dots$
Q11 — Combination of Resistors · easy · numerical
Two resistors are connected (a) in series (b) in parallel. The equivalent resistance in the two cases are 9 ohm and 2 ohm respectively. Then the resistances of the component resistors are
A. 3 ohm and 6 ohm ✓ Correct
B. 2 ohm and 7 ohm
C. 3 ohm and 9 ohm
D. 5 ohm and 4 ohm
Solution: $R_1 + R_2 = 9$ and $R_1R_2/9 = 2$ give $R_1R_2 = 18$. The numbers with sum 9 and product 18 are 3 and 6.
Q12 — Combination of Resistors · easy · numerical
Resistors of 1, 2, 3 ohm are connected in the form of a triangle. If a 1.5 volt cell of negligible internal resistance is connected across 3 ohm resistor, the current flowing through this resistance will be
A. 0.5 amp ✓ Correct
B. 0.25 amp
C. 1.0 amp
D. 1.5 amp
Solution: The cell is directly across the 3 $\Omega$ resistor (internal resistance negligible), so the current through it is $i = V/R = 1.5/3 = 0.5\ A$.
Q13 — Combination of Resistors · easy · theory
A student has 10 resistors of resistance '$r$'. The minimum resistance made by him from given resistors is
A. $10r$
B. $\dfrac{r}{100}$
C. $\dfrac{r}{10}$ ✓ Correct
D. $\dfrac{r}{5}$
Solution: The resistance is least when all ten are joined in parallel, giving $R_{eq} = r/10$.
Q14 — Combination of Resistors · easy · numerical
A copper wire of resistance $R$ is cut into ten parts of equal length. Two pieces each are joined in series and then five such combinations are joined in parallel. The new combination will have a resistance
A. $\dfrac{R}{25}$ ✓ Correct
B. $\dfrac{R}{4}$
C. $R$
D. $\dfrac{R}{5}$
Solution: Each piece is $R/10$; two in series give $R/5$; five such in parallel give $(R/5)/5 = R/25$.
Q15 — Combination of Resistors · easy · numerical
A wire has resistance $12\ \Omega$. It is bent in the form of a circle. The effective resistance between the two points on any diameter is equal to
A. $6\ \Omega$
B. $12\ \Omega$
C. $24\ \Omega$
D. $3\ \Omega$ ✓ Correct
Solution: A diameter splits the ring into two halves of $6\ \Omega$ each, and they are in parallel: $6 \times 6/12 = 3\ \Omega$.
Q16 — Combination of Resistors · easy · numerical
Three resistances $4\ \Omega$ each of are connected in the form of an equilateral triangle. The effective resistance between two corners is
A. $\dfrac{3}{8}\ \Omega$
B. $\dfrac{8}{3}\ \Omega$ ✓ Correct
C. $12\ \Omega$
D. $8\ \Omega$
Solution: Between two corners, one side ($4\ \Omega$) is in parallel with the other two in series ($8\ \Omega$): $4 \times 8/12 = 8/3\ \Omega$.
Q17 — Combination of Resistors · easy · numerical
10 wires (same length, same area, same material) are connected in parallel and each has $1\ \Omega$ resistance, then the equivalent resistance will be
A. $1\ \Omega$
B. $10\ \Omega$
C. $0.001\ \Omega$
D. $0.1\ \Omega$ ✓ Correct
Solution: For $n$ equal resistances in parallel $R_{eq} = R/n = 1/10 = 0.1\ \Omega$.
Q18 — Combination of Resistors · easy · numerical
Four resistances $10\ \Omega$, $5\ \Omega$, $7\ \Omega$ and $3\ \Omega$ are connected so that they form the sides of a rectangle $AB$, $BC$, $CD$ and $DA$ respectively. Another resistance of $10\ \Omega$ is connected across the diagonal $AC$. The equivalent resistance between $A$ and $B$ is
A. $7\ \Omega$
B. $10\ \Omega$
C. $2\ \Omega$
D. $5\ \Omega$ ✓ Correct
Solution: Between $A$ and $C$, the diagonal ($10\ \Omega$) is in parallel with $AD + DC = 3 + 7 = 10\ \Omega$, giving $5\ \Omega$. Adding $CB = 5\ \Omega$ gives $10\ \Omega$, which is in parallel with $AB = 10\ \Omega$: $R_{AB} = 5\ \Omega$.
Q19 — Combination of Resistors · easy · theory
Two wires of equal diameters, of resistivities $\rho_1$ and $\rho_2$ and lengths $l_1$ and $l_2$, respectively, are joined in series. The equivalent resistivity of the combination is
A. $\dfrac{\rho_1 l_1 + \rho_2 l_2}{l_1 + l_2}$ ✓ Correct
B. $\dfrac{\rho_1 l_2 + \rho_2 l_1}{l_1 - l_2}$
C. $\dfrac{\rho_1 l_2 + \rho_2 l_1}{l_1 + l_2}$
D. $\dfrac{\rho_1 l_1 - \rho_2 l_2}{l_1 - l_2}$
Solution: In series $R = R_1 + R_2$, so $\dfrac{\rho(l_1+l_2)}{A} = \dfrac{\rho_1 l_1}{A} + \dfrac{\rho_2 l_2}{A}$, giving $\rho = \dfrac{\rho_1 l_1 + \rho_2 l_2}{l_1 + l_2}$.
Q20 — Combination of Resistors · easy · numerical
Four resistances of $100\ \Omega$ each are connected in the form of square. Then, the effective resistance along the diagonal points is
A. $100\ \Omega$ ✓ Correct
B. $200\ \Omega$
C. $150\ \Omega$
D. $400\ \Omega$
Solution: Along a diagonal, two arms of $200\ \Omega$ each are in parallel: $200 \times 200/400 = 100\ \Omega$.
Q21 — Combination of Resistors · easy · numerical
Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other and the thinner wire has a resistance of 8 ohms, the resistance of the combination is equal to
A. $\dfrac{3}{8}$ ohms
B. $\dfrac{5}{8}$ ohms
C. $\dfrac{8}{3}$ ohms
D. $\dfrac{8}{5}$ ohms ✓ Correct
Solution: The thicker wire has twice the diameter, hence four times the area and a quarter of the resistance: $8/4 = 2\ \Omega$. In parallel, $8 \times 2/10 = 8/5\ \Omega$.
Q22 — Combination of Resistors · easy · numerical
Two resistance wires on joining in parallel the resultant resistance is $\dfrac{6}{5}$ ohms. One of the wire breaks, the effective resistance is 2 ohms. The resistance of the broken wire is
A. 3 ohm ✓ Correct
B. $\dfrac{3}{5}$ ohm
C. $\dfrac{6}{5}$ ohm
D. 2 ohm
Solution: The unbroken wire is $2\ \Omega$. From $\dfrac{5}{6} = \dfrac{1}{2} + \dfrac{1}{x}$ we get $\dfrac{1}{x} = \dfrac{1}{3}$, so the broken wire is $3\ \Omega$.
Q23 — Combination of Resistors · easy · numerical
Three resistors are connected to form the sides of a triangle ABC, the resistance of the sides AB, BC and CA are 40 ohms, 60 ohms and 100 ohms respectively. The effective resistance between the points A and B in ohms will be
A. 50
B. 200
C. 32 ✓ Correct
D. 64
Solution: Between $A$ and $B$, the side $AB$ ($40\ \Omega$) is in parallel with $BC + CA = 160\ \Omega$: $40 \times 160/200 = 32\ \Omega$.
Q24 — Combination of Resistors · easy · theory
Two wires of the same dimensions but resistivities $\rho_1$ and $\rho_2$ are connected in series. The equivalent resistivity of the combination is
A. $\dfrac{\rho_1 + \rho_2}{2}$ ✓ Correct
B. $\rho_1 + \rho_2$
C. $\sqrt{\rho_1\rho_2}$
D. $2(\rho_1 + \rho_2)$
Solution: With equal lengths and areas, $\dfrac{\rho(2l)}{A} = \dfrac{\rho_1 l}{A} + \dfrac{\rho_2 l}{A}$, so $\rho = \dfrac{\rho_1 + \rho_2}{2}$.
Q25 — Combination of Resistors · easy · numerical
Three unequal resistors in parallel are equivalent to a resistance 1 ohm. If two of them are in the ratio $1 : 2$ and if no resistance value is fractional, the largest of the three resistances in ohms is
A. 4
B. 8
C. 6 ✓ Correct
D. 12
Solution: Take the resistors as $x$, $2x$ and $y$. With $x = 3$: $\dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{2}$, so $y = 2$. The three are 3, 6 and 2 $\Omega$, and the largest is $6\ \Omega$.
Q26 — Combination of Resistors · easy · numerical
An unknown resistance $R_1$ is connected in series with a resistance of $10\ \Omega$. This combinations is connected to one gap of a metre bridge while a resistance $R_2$ is connected in the other gap. The balance point is at 50 cm. Now, when the $10\ \Omega$ resistance is removed the balance point shifts to 40 cm. The value of $R_1$ is (in ohm)
A. 60
B. 10
C. 40
D. 20 ✓ Correct
Solution: At 50 cm, $R_1 + 10 = R_2$. After removing the $10\ \Omega$, $\dfrac{R_1}{R_2} = \dfrac{40}{60} = \dfrac{2}{3}$. Substituting gives $3R_1 = 2R_1 + 20$, so $R_1 = 20\ \Omega$.
Q27 — Combination of Resistors · easy · numerical
An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 4/3 and 2/3, then the ratio of the currents passing through the wire will be
A. 3
B. 2
C. 1/3 ✓ Correct
D. 8/9
Solution: In parallel the currents divide in the ratio of the conductances, $i \propto A/l \propto r^{2}/l$. So $\dfrac{i_1}{i_2} = \left(\dfrac{2}{3}\right)^{2} \times \dfrac{3}{4} = \dfrac{1}{3}$.
Q28 — Combination of Resistors · easy · theory
If a rod has resistance $4\ \Omega$ and if rod is turned as half cycle then the resistance along diameter
A. $1.56\ \Omega$
B. $4\ \Omega$ ✓ Correct
C. $2.44\ \Omega$
D. $2\ \Omega$
Solution: Bending the rod into a semicircle leaves the whole rod as the only conducting path between the two ends of the diameter, so the resistance stays $4\ \Omega$.
Q29 — Combination of Resistors · easy · numerical
A current of 2 A flows in a system of conductors as shown. The potential difference $(V_A - V_B)$ will be
A. $-2\ V$
B. $+1\ V$ ✓ Correct
C. $+2\ V$
D. $-1\ V$
Solution: Both branches from $D$ to $C$ measure $2 + 3 = 5\ \Omega$, so the 2 A splits into 1 A each. Then $V_D - V_A = 1 \times 2 = 2\ V$ and $V_D - V_B = 1 \times 3 = 3\ V$, giving $V_A - V_B = +1\ V$.
Q30 — Combination of Resistors · easy · numerical
Resistances of 6 ohm each are connected in the manner shown in adjoining figure. With the current 0.5 ampere as shown in figure, the potential difference $V_P - V_Q$ is
A. 7.2 V
B. 3.6 V
C. 3.0 V ✓ Correct
D. 6.0 V
Solution: The left group is three $6\ \Omega$ in parallel $= 2\ \Omega$; the right group is $(6 + 6) = 12\ \Omega$ in parallel with $6\ \Omega = 4\ \Omega$. Total $= 6\ \Omega$, so $V_P - V_Q = 0.5 \times 6 = 3.0\ V$.