Current Electricity - NEET Physics Chapterwise MCQs & PYQs

NEET Current Electricity MCQs & PYQs

Practice NEET Current Electricity Questions

Question 191:

moderate

You are given several identical resistances each of value $R = 10 \Omega$ and each capable of carrying a maximum current of one ampere. It is require to make a suitable combination of these resistances of $5 \Omega$ which can carry a current of $4 \text{ ampere}$. The minimum number of resistances of the type $R$ that will be required for this job is:

(1990)

To carry $4 \text{ A}$ safely, 4 parallel branches are needed (each carries $1 \text{ A}$). Let each branch have resistance $r$. Equivalent resistance $r/4 = 5 \Omega$, so $r = 20 \Omega$. Each branch needs two $10 \Omega$ resistors in series. Total resistors = $4 \times 2 = 8$.

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Question 192:

easy

A resistance wire connected in the left gap of a metre bridge balances a $10\Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3 : 2$. If the length of the resistance wire is $1.5\text{ m}$, then the length of $1\Omega$ of the resistance wire is:

(2020)

Let the resistance of the left gap be $R$. For a balanced metre bridge, $\frac{R}{10} = \frac{3}{2} \implies R = 15\Omega$.nThe length of the $15\Omega$ resistance wire is $1.5\text{ m}$.nTherefore, the length of $1\Omega$ of the wire is $\frac{1.5}{15} = 0.1\text{ m} = 1.0 \times 10^{-1}\text{ m}$.

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Question 193:

easy

The resistances of the four arms P, Q, R and S in a Wheatstone’s bridge are $10\text{ ohm}$, $30\text{ ohm}$, $30\text{ ohm}$ and $90\text{ ohm}$, respectively. The e.m.f. and internal resistance of the cell are $7\text{ volt}$ and $5\text{ ohm}$ respectively. If the galvanometer resistance is $50\text{ ohm}$, the current drawn from the cell will be:

(2013)

The bridge is balanced because $\frac{P}{Q} = \frac{10}{30} = \frac{1}{3}$ and $\frac{R}{S} = \frac{30}{90} = \frac{1}{3}$. No current flows through the galvanometer.nEquivalent resistance of the bridge $R_{eq} = \frac{(10+30) \times (30+90)}{(10+30) + (30+90)} = \frac{40 \times 120}{160} = 30\Omega$.nTotal resistance $= 30\Omega + 5\Omega = 35\Omega$. Current $I = \frac{V}{R_{total}} = \frac{7}{35} = 0.2\text{ A}$.

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Question 194:

easy

Three resistances P, Q, R each of $2\Omega$ and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of $6\Omega$ is connected in parallel to S the bridge gets balanced. What is the value of S?

(2007)

For the Wheatstone bridge to be balanced with $P=Q=R=2\Omega$, the equivalent resistance of the fourth arm must also be $2\Omega$.nThe fourth arm is $S$ in parallel with $6\Omega$, so $\frac{6S}{S + 6} = 2$.n$6S = 2S + 12 \implies 4S = 12 \implies S = 3\Omega$.

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Question 195:

easy

A $6\text{ volt}$ battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of $100\text{ ohm}$. The difference of potential between two points on the wire separated by a distance of $50\text{ cm}$ will be:

(2004)

The potential gradient $k = \frac{V}{L} = \frac{6\text{ V}}{3\text{ m}} = 2\text{ V/m}$.nThe potential difference across a $50\text{ cm}$ ($0.5\text{ m}$) segment is $\Delta V = k \times l$.n$\Delta V = 2\text{ V/m} \times 0.5\text{ m} = 1\text{ V}$.

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Question 196:

easy

In a Wheatstone’s bridge all the four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is:

(2003)

Since all four arms have resistance $R$, the ratio of adjacent arms is equal ($R/R = R/R$), making it a balanced Wheatstone bridge.nNo current flows through the galvanometer arm, so it can be ignored.nEquivalent resistance $R_{eq} = \frac{(R+R)(R+R)}{(R+R)+(R+R)} = \frac{2R \times 2R}{4R} = R$.

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Question 197:

easy

The resistance of each arm of the Wheatstone bridge is $10\text{ ohm}$. A resistance of $10\text{ ohm}$ is connected in series with galvanometer then the equivalent resistance across the battery will be:

(2001)

The bridge has equal resistance ($10\text{ ohm}$) in all four arms, so it is balanced.nBecause it is balanced, the current through the galvanometer branch is zero, making its total resistance irrelevant.nThe equivalent resistance of the circuit is $\frac{(10+10) \times (10+10)}{(10+10) + (10+10)} = 10\text{ ohm}$.

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Question 198:

easy

In a potentiometer circuit a cell of EMF $1.5\text{ V}$ gives balance point at $36\text{ cm}$ length of wire. If another cell of EMF $2.5\text{ V}$ replaces the first cell, then at what length of the wire, the balance point occurs?

(2021)

For a potentiometer, the EMF is directly proportional to the balancing length: $\frac{E_1}{E_2} = \frac{l_1}{l_2}$.nSubstitute the given values: $\frac{1.5}{2.5} = \frac{36}{l_2}$.n$l_2 = 36 \times \frac{2.5}{1.5} = 36 \times \frac{5}{3} = 60\text{ cm}$.

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Question 199:

easy

A potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given by:

(2015 Re)

Current through the potentiometer wire is $I = \frac{E_0}{r + r_1}$.
Potential gradient is $k = \frac{I r}{L} = \frac{E_0 r}{(r + r_1) L}$.
The unknown e.m.f. $E = k \cdot l = \frac{E_0 r}{(r + r_1)} \frac{l}{L}$.

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Question 200:

easy

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of $2.0\text{ V}$ and a negligible internal resistance. The potentiometer wire itself is $4\text{ m}$ long. When the resistance, $R$, connected across the given cell, has values of (i) Infinity, (ii) $9.5\text{ }\Omega$
The ‘balancing lengths’, on the potentiometer wire are found to be $3\text{ m}$ and $2.85\text{ m}$, respectively. The value of internal resistance of the cell is:

(2014)

The internal resistance formula for a potentiometer is $r = R \left(\frac{l_1}{l_2} - 1\right)$.
Substituting $l_1 = 3\text{ m}$, $l_2 = 2.85\text{ m}$, and $R = 9.5\text{ }\Omega$:
$r = 9.5 \cdot \left(\frac{3}{2.85} - 1\right) = 9.5 \cdot \frac{0.15}{2.85} = 0.5\text{ }\Omega$.

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