Name: RC Circuit and Charging Capacitor
Uploaded: 2019-07-19
Duration: 13 min 36 s
Description: In this video, we will discuss the physics of charging the capacitor connected in a circuit and derive the expression and relation between the parameters inv...

Question

Two resistors,

400 Ω,

and

800 Ω

are connected in series with a

6 V

battery. It is desired to measure the current in the circuit. An ammeter of

10 Ω

resistance is used for this purpose. The reading of ammeter will be

\frac{N}{1210} A .

Similarly, if a voltmeter of

1000 Ω

resistance is used to measure the potential difference across the

400 Ω

resistor, the reading of voltmeter is

\frac{P}{19} V .

Then the value of

N

and

P

are:

Accepted Answer

Diagram for the first case when ammeter is connected to measure current is shown below.

Since, ammeter has a resistance of $10 Ω$ and is connected in series with the other two resistors. So, the equivalent resistance in the circuit will be,

$R_{eq} = 400 + 800 + 10 = 1210 Ω$ ,

Given supply voltage is, $V = 6 V$ ,

So, current in the circuit, $400 Ω$ is;

$I = \frac{V}{R} = \frac{6}{1210} A$ ,

Now, comparing the calculated current with the ammeter reading, we get $N = 6$ .

Again, the circuit diagram to measure voltage across the given resistor is shown below;

Since, voltmeter has a resistance of $1000 Ω$ and is parallel with $400 Ω$ resistor and their combination is in series with $800 Ω$ resistor, then,

The equivalent resistance in the circuit is given as;

$R_{eq} = 800 + (\frac{1000 \times 400}{1000 + 400}) = 800 + \frac{2000}{7}$

$\Rightarrow R_{eq} = \frac{7600}{7} Ω$

Now, current in the circuit is;

$I = \frac{V}{R_{eq}} = \frac{6}{\frac{7600}{7}} = \frac{42}{7600} A,$

Then, potential difference across resistor $400 Ω$ is;

$V = I R = \frac{42}{7600} \times \frac{2000}{7} = \frac{30}{19}$

Now, comparing the calculated voltage and given reading we get,

$P = 30$ .

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