Question

A circuit contains an ammeter, a battery of 30V and a resistance of 40.8 ohm all connected series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading ammeter will be?

Answer 1

Hint: For answering these types of questions we must remember the concept of resistance connected in series and in parallel. We must know how to calculate the effective resistance in both types of arrangements.
In a huge and complicated circuit with many resistors, the effective resistance is the total resistance of the circuit which is usually measured between 2 points.

Complete step-by-step answer:
The two types of arrangements are

(i)Series connection
(ii)Parallel connection

If individual resistors are connected from end to end then resistors are said to be connected in series.

The effective resistance of three resistors of resistances ${R_1},{R_2} and\;{R_3}$ connected in series is given by

${R_{effective}} = {R_1} + {R_2} + {R_3} + .....$

If each end of individual resistors is connected together to one another as one, then resistors are said to be connected in parallel.

The effective resistance, R, of three resistors of resistances ${R_1},{R_2} and\;{R_3}$ connected in parallel is given by:

$\dfrac{1}{{{R_{effective}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .....$
${R_{effective}} = 40.8 + \dfrac{{480 \times 20}}{{480 + 20}}$
${R_{effective}} = 40.8 + 19.2$
${R_{effective}} = 60\;\Omega $

So, the current flowing across ammeter

$I = \dfrac{V}{R} = \dfrac{{30}}{{60}} = \dfrac{1}{2} = 0.5A$

Therefore, the effective current in the circuit is 0.5 Ampere and the reading ammeter will be 0.5 A

Note: In a series connection, the effective resistance, ${R_{effective}}$ is always larger than the largest of the individual resistances and in a parallel connection, the effective resistance, ${R_{effective}}$ is always smaller than the smallest of the individual resistances.