Question

A galvanometer having a coil resistance $60$ shows a full-scale deflection when a current of $1.0\,A$ passes through it. It can be converted into an ammeter to read current upto $5.0\,A$ by
A. putting in series a resistance of $15$
B. putting in series a resistance of $240$
C. putting in parallel a resistance of $15$
D. putting in parallel a resistance of $240$

Answer 1

We can start by drawing a diagram to represent the question. We can then move onto finding the voltage associated with the current and resistance when another resistance is not connected. After we have done this, we can take the value of total current as the sum of series current and the current through the added resistance. We can equate it to the voltage which does not change here and get the value of resistance. This will lead us to the required solution

Formulas used:
The formula relating the voltage, current and resistance is given by the ohm's law and it is as,
$V = IR$
The total current through the circuit is given by the formula,
$I = {I_s} + {I_{new}}$
Where ${I_s}$ is the series current or the current through the circuit before the addition of the new resistance and ${I_{new}}$ is the current through the new added resistance.

Complete step by step answer:
Let us start by noting down the quantities given in the question. The current through the circuit before the addition of the new resistance or the series current is ${I_s} = 1A$.The resistance before the new resistance is added or the series resistance is given as ${R_s} = 60\Omega$.The total current is given as $I = 5A$

We can find the voltage using the ohms law as,

\Rightarrow V = 1 \times 60 \\\ \Rightarrow V = 60\,V$$ The voltage will always be constant and hence we can equate the value we got now with that voltage through the new resistance which is $$V = {I_{new}}{R_{new}}$$ In order to get the value of current, we use the formula for the total energy which is, $$I = {I_s} + {I_{new}}$$ We can rearrange and get $${I_{new}} = I - {I_s}$$. That is $${I_{new}} = I - {I_s} \\\ \Rightarrow 5 - 1 = 4A$$ By equating it with the initial value, we have $${I_{new}}{R_{new}} = {I_s}{R_s}$$ We find the value of new resistance as $${R_{new}} = \dfrac{{{I_s}{R_s}}}{{{I_{new}}}} \\\ \Rightarrow {R_{new}} = \dfrac{{1 \times 60}}{4} \\\ \therefore {R_{new}} = 15\Omega $$ Now we have to find whether this is in series or in parallel. Since the current is increasing, the resistance must decrease. Hence, the connection must be in parallel. **Hence, the correct answer is option C.** **Note:** We took the assumption that the resistance will be parallel because, when resistors are connected in series, the total resistance will be bigger than both the values of resistance. If the resistors are connected in parallel, the value of total resistance will be smaller than the smallest resistor among the individual resistors.