Question

A potential difference of 220 V is maintained across a 12000 ohm rheostat, as shown in the figure. The voltmeter has a resistance of 6000 ohm and point c is at one-fourth of the distance from a to b, therefore, the reading of the voltmeter will be

A. 32 V
B. 36 V
C. 40 V
D. 42 V

Answer 1

Hint:This problem is from the section current electricity. The concept of equivalent resistance is needed to solve this problem. The current decreases as resistance increases. On the other hand, the current increases as the resistance decreases.

Formula used:
Equivalent resistance for a series resistance circuit:
${R_s} = {R_1} + {R_2} + {R_3}$
Where ${R_s}$= series equivalent resistance and ${R_1},{R_2},{R_3}$ = component resistance.
Equivalent resistance for a parallel resistance circuit:
$\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}$
Where ${R_p}$= parallel equivalent resistance and ${R_1},{R_2},{R_3}$ = component resistance.

Complete step by step solution:
In the case of linear Rheostat, Resistance (R) is directly proportional to length (L).
$R \propto L$
Then, ${R_{ac}} \propto ac$ and ${R_{ab}} \propto ab$.
Taking the ratio in the values and we will get
$\dfrac{{{R_{ac}}}}{{{R_{ab}}}} = \dfrac{{ac}}{{ab}}$
The given value is:
${R_{ab}} = 12000\Omega $
And, $ac = \dfrac{1}{4}ab \Rightarrow \dfrac{{ac}}{{ab}} = \dfrac{1}{4}$
Substitute these values and find ${R_{ac}}$.
${R_{ac}} = 12000 \times \dfrac{1}{4} = 3000\Omega $
The resistance${R_{ac}}$ is in parallel with the voltmeter. The resistance of the voltmeter is given and that is ${V_R} = 6000\Omega $.

The resistances ${R_{ac}}$ and ${V_R}$ are in parallel connection. Then, the effective resistance between points a and c is,
${R'_{ac}} = \dfrac{{{R_{ac}} \times {V_R}}}{{{R_{ac}} + {V_R}}} \\
\Rightarrow {R'_{ac}} = \dfrac{{3000 \times 6000}}{{9000}} \\
\Rightarrow {R'_{ac}} = 2000\Omega \\ $
The resistance between points a and c is,
${R_{bc}} = 12000 - 3000 = 9000\Omega $.
${R_{bc}}$ and ${R'_{ac}}$ care in series.
Therefore the Voltmeter reading will be,
${V_{ac}} = \dfrac{{{{R'}_{ac}} \times {V_{ab}}}}{{{{R'}_{ac}} + {R_{bc}}}} \\
\therefore {V_{ac}} = \dfrac{{2000 \times 220}}{{9000 + 2000}} = 40V$

Hence, the correct option is option C.

Note: A conductor's electrical resistance is affected by the following parameters: The conductor's cross-sectional area, the conductor's length, the conductor's material and the conducting material's temperature. Electrical resistance is inversely proportional to the cross-sectional area and directly proportional to the conductor's length.