Question

In the circuit shown, resistance of galvanomete
is 100 2. It is being used as a voltmeter. Its reading
will be
500
1000
500
10V
(1) O volt
(2) 2 volt
(3) 5 volt
(4) 10 volt​

Answer 1

Answer:

5V

Explanation:

Total Reistance = 200 ohm

Current = V/R = 10/200 = 0.05 A

Galvanometer will show the reading = potential difference across it (= potential difference across 100 ohm)

Potential difference across 100 ohm = IR

= 0.05 x 100 = 5 V

Answer 2

The reading of the voltmeter will be (3) 5 volt

In this circuit, as there is one wire connected parallel to the first 50Ω resistance, it will be neglected.

The resistance of the galvanometer = 100 Ω

So, the equivalent resistance of the circuit = (100/2)+50 Ω

                                                                         = 100 Ω

The voltage of the circuit = 10 V

∴ The current flowing through the whole circuit (I) = 10/100 A

                                                                               = 0.1 A

Now, the Galvanometer will show the voltage, V = IR

The equivalent resistance across the galvanometer (R) = 100/2 Ω

                                                                                      = 50 Ω

∴ The voltage across the galvanometer = 0.1 × 50 V

                                                                  = 5 V