In the given circuit, a galvanometer with a resistance of `70 Omega` is converted to an ammeter `0.05 Omega` by a shunt resistance of 0.05 2, total current measured by this device

Options

A. 0.88A
B. 0.77A
C. 0.55A
D. 0.99A

Answer

D

Explanation

The equivalent circuit of given circuit is given by
Here, `R_(S) = (R_(G)r_(s))/(R_(G) + r_(s)) = (70 xx 0.05)/(70+0.05)`
`therefore R_(s) = 0.0499 Omega = 0.05 Omega`
The total resistance in the circuit
`R=R_(s) + 5Omega = 0.05 + 5 = 5.05 Omega`
The current measured by the device
`I=V/R = 5/(5.05) = 0.99 A`