Question

The internal resistance of voltmeter V is 5000Ω. If it's connected:-​

Answer 1

Answer:

(4) is the correct option

Explanation:

First let us find the effective resistance of the given circuit

Voltmeter has 5000 ohm resistance which is parallel to 500 ohm

Now, effective resistance is

$\mathrm{R_p = \dfrac{5000.500}{5000+500}}$

$\implies \mathrm{R_p = \dfrac{5000.5\not 0\not0}{55\not0\not0}}$

$\implies \mathrm{R_p = \dfrac{5000\times 5}{55}}$

$\implies \mathrm{R_p = \dfrac{5000}{11}}$

$\mathrm{R_p = 454.54}$

So, overall R = 100 + 454.54 = 554.54

Now, current through circuit is

$\boxed{\mathrm{I = \dfrac{V}{R}}}$

$\mathrm{I = \dfrac{5.54}{554}}$

I = 0.01 A

So,

now let us find the potential difference at the resistors

Potential difference across R1 is

$\mathrm{V_1 = I \times R_1}$

$\mathrm{V_1 = 0.01 \times 100}$

$\mathrm{V_1 = 1V}$

Now, V2 = 5.54 - 1 = 4.54

So, Across R2 , V = 4.54V

So, both the options (1) and (2) are wrong.

• Now, going to the options (3) and (4),

It is asked that what will be the voltage shown in the voltmeter if it is connected in parallel with both R1 and R2

We know that,

the voltage is same at parallel

So, in parallel, Voltmeter shows 5.54V

Next, It is asked that what will be the voltage shown in the voltmeter if it is connected in series with both R1 and R2

As it is in series, Voltage will be less than 5.54 as in series voltage is distributed to each resistor

So, It will read more in parallel than in series.

Thus, (4) is the correct option

Hope it helps!!