Question 11:
easyAssertion (A): In a balanced Wheatstone bridge, the current through cell depends on resistance of galvanometer.
Reason (R): At balanced condition current through galvanometer is non-zero.
In a balanced Wheatstone bridge, the current through the galvanometer is zero. Hence, the galvanometer's resistance does not affect the equivalent resistance of the bridge and thus the current drawn from the cell. Both Assertion (A) and Reason (R) are false.
Question 12:
easyA circuit contains an ammeter, a battery of $30 \text{ V}$ and a resistance $40.8 \Omega$ all connected in series. If the ammeter has a coil of resistance $480 \Omega$ and a shunt of $20 \Omega$, the reading in the ammeter will be:
(2015 Re)
Total resistance of ammeter $R_A = \frac{480 \times 20}{480 + 20} = 19.2 \Omega$. Total resistance of circuit $R_{eq} = 40.8 + 19.2 = 60 \Omega$. Reading of ammeter $I = \frac{V}{R_{eq}} = \frac{30}{60} = 0.5 \text{ A}$.
Question 13:
easyA $6\text{ volt}$ 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$\Delta V = 2\text{ V/m} \times 0.5\text{ m} = 1\text{ V}$.
Question 14:
easyIn 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 $R_{eq} = \frac{(R+R)(R+R)}{(R+R)+(R+R)} = \frac{2R \times 2R}{4R} = R$.
Question 15:
easyThe 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 $\frac{(10+10) \times (10+10)}{(10+10) + (10+10)} = 10\text{ ohm}$.
Question 16:
easyIn 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$l_2 = 36 \times \frac{2.5}{1.5} = 36 \times \frac{5}{3} = 60\text{ cm}$.
Question 17:
easyA 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 $\frac{1.5}{15} = 0.1\text{ m} = 1.0 \times 10^{-1}\text{ m}$.
Question 18:
easyThe 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 $I = \frac{V}{R_{total}} = \frac{7}{35} = 0.2\text{ A}$.
Question 19:
easyThree 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$6S = 2S + 12 \implies 4S = 12 \implies S = 3\Omega$.
Question 20:
easyA potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves:
(2017-Delhi)
A potentiometer measures the electromotive force (e.m.f.) of a cell using a null deflection method.
At the balance point, no current flows through the galvanometer, so no current is drawn from the cell being measured.
This allows it to measure the true open-circuit potential difference accurately.