Question

A voltmeter V of 23KΩ resistance in series with a resistor R across a 230 V supply reads 92 Volts. Calculate the value of the resistor R?​

Answer 1

Answer:

Given :

Resistance of Voltmeter=R=RV=23k=23k\OmegaΩ

Reading of voltmeter = Potential difference across the voltmeter =92V=92V

External source is applied of voltage =230V=230V

So Potential difference across resistance r will be = 230V-92V= 138V=230V−92V=138V

To calculate the value of resistance r

from Diagram

Current through the voltmeter and resistance r will be

I= \frac{230V}{R_v + r}I=

R

v

+r

230V

by Ohm's Law

Now potential difference across voltmeter = reading of voltmeter =92V=92V = IR_v=IR

v

= \frac{230R_V}{R_v+r}=

R

v

+r

230R

V

after putting value of RRv

\implies⟹ \frac{r}{R_v} = \frac{230}{92}-1

R

v

r

=

92

230

−1

\implies r = 1.5 \times R_v⟹r=1.5×R

v

= 34500\Omega34500Ω

Final answer

The value of resistance r connected in series with voltmeter is 34.5k\Omega34.5kΩ .

Answer 2

Answer:

34.5kΩ

Explanation:

Given that:

Voltmeter of resistance in series = 23KΩ

Resistance = 230 V

Supply = 92 V

To calculate:

Value of resistor R =?

Solution:

Let $R_{v}$ = 23kΩ

External sources of voltage applied = 230V

Formula for voltmeter:

Reading of voltmeter = potential difference between the voltmeter will be 92V

Hence, the potential difference of resistance = 230V – 92V =138V

Let us consider that,

The current passing through voltmeter and resistance r

I = $230V$/$R_{v} + r$ (by using the ohm’s law)

Let the potential difference across voltmeter = reading of voltmeter

= 92V = I $R_{v}$ = $230R_{v}$/ $R_{v} +r$

Now,

$r/ R_{v}$ = 230/92-1

r = 1.5 x RV = 34500Ω

Therefore, form the above calculation the resistance r is connected in series with voltmeter will be 34.5Ω.

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