Question

An ammeter with a scale ranging from $0$ to $15mA$ has a resistance of $5\Omega $ . How should the instrument be connected with a resistor (and with what resistance) to measure: (a) current of $0$ to $0.15A$, (b) a potential difference from $0$ to $150V$?

Answer 1

Hint:To solve this question, we need to understand the connection of resistors in which voltage is constant or current is constant. In the first case, we have to find the resistor through which we can change the range of the current. Whereas in the second case, we have to find the resistor through which we can change the range of the voltage. For this, we will use the concept of parallel and series connection of resistors.

Complete step by step answer:
In case a, we need to find the type of connection and amount of resistance to be connected with the instrument to measure a current of $0$to $0.15A$. We know that the variation in current is possible when the voltage remains constant. This is possible if the resistor is connected in parallel with the instruments.So let us assume that a resistor R is connected in parallel with the instrument. We will take \[{I_1} = 15mA = 0.015A\], ${R_1} = 5\Omega $, ${I_2} = 0.15 - 0.015 = 0.135A$ and ${R_2} = R\Omega $
For parallel connection of resistors,
$
{I_1}{R_1} = {I_2}{R_2} \\
\Rightarrow 0.015 \times 5 = 0.135 \times R \\
\therefore R = \dfrac{5}{9}\Omega \\ $
Thus, our answer for case A is: The resistance of $\dfrac{5}{9}\Omega $ should be connected in parallel with the instrument.

Now, in case b, we need to find the type of connection and amount of resistance to be connected with the instrument to measure a potential difference from $0$to $150V$.
The variation in voltage will be done if the resistor is connected in series.
Let us take $V = 150V$, $I = 15mA = 0.015A$, ${R_1} = 5\Omega $ and ${R_2} = R\Omega $
We know that for series connection,
\[
V = I\left( {{R_1} + {R_2}} \right) \\
\Rightarrow 150 = 0.015\left( {5 + R} \right) \\
\Rightarrow R + 5 = 10000 \\
\Rightarrow R = 10000 - 5 \\
\therefore R = 9995\Omega \\ \]
Thus, our answer for case B is: The resistance of $9995\Omega $ should be connected in series with the instrument.

Note:We have used two important concepts. First, we have applied the fact that in the parallel connection, the total current will divide between two resistors and the potential difference is constant. Second, in the series connection of the resistors, the current remains constant in both the resistors and potential difference is divided between the resistors.