Name: RC Circuit and Charging Capacitor
Uploaded: 2019-07-19
Duration: 13 min 36 s
Description: In this video, we will discuss the physics of charging the capacitor connected in a circuit and derive the expression and relation between the parameters inv...

Question

A galvanometer (coil resistance $99 Ω$ ) is converted into a ammeter using a shunt of $1 Ω$ and connected as shown in the figure (i). The ammeter reads $3 A$ . The same galvanometer is converted into a voltmeter by connected a resistance of $101 Ω$ in series. This voltmeter is connected as shown in figure (ii). Its reading is found to be $\frac{4}{5}$ of the full scale reading. Find range of the ammeter and voltmeter.

Accepted Answer

Given,
Galvanometer's coil resistance is, $G = 99 Ω$
Shunt resistance in parallel is, $s = 1 Ω$
Reading of ammeter is, $i = 3 A$
Shunt resistance of voltmeter is, $S = 101 Ω$

Resistance of ammeter is given by, $R_{A} = \frac{s \times G}{s + G}$
$\Rightarrow R_{A} = \frac{99 \times 1}{100} = 0.99 Ω$

Applying Ohm's law in first circuit, we have
$\Rightarrow 3 (2 + r + 0.99) = 12$
$\Rightarrow 2 + r + 0.99 = 4$

$\Rightarrow r = 1.01 Ω$

Using second circuit, we have

Resistance of voltmeter is, $R_{V} = G + S$
$\Rightarrow R_{V} = 99 + 101 = 200 Ω$

Net resistance of the circuit is,
$R_{e q} = \frac{1}{\frac{1}{R_{V}} + \frac{1}{2}} + r$
$\Rightarrow R_{e q} = \frac{1}{\frac{1}{200} + \frac{1}{2}} + 1.01 \approx 3 Ω$

Current in the circuit is, $i' = \frac{V}{R_{e q}}$
$\Rightarrow i' = \frac{12}{3} = 4 A$

In the given circuit, voltage across voltmeter will be equal to terminal voltage across battery.
$V_{V} = V - i' r$
$\Rightarrow V_{V} = 12 - 4 \times 1.01 = 7.96 V$

According to question,
$V_{V} = \frac{4}{5} V_{V, m a x}$
$\Rightarrow V_{V, m a x} = \frac{5}{4} V_{V}$
$\Rightarrow V_{V, m a x} = \frac{5}{4} \times 7.96 = 9.95 V$

Maximum galvanometer current is, $i_{g} = \frac{V_{V, m a x}}{R_{V}} = \frac{9.95}{200} \approx 0.05 A$

Range of ammeter and voltmeter is to be calculated using this maximum galvanometer current.
In case of ammeter, we have
$i_{g} \times G = (i - i_{g}) s$ , here $i$ represents maximum current through ammeter.
$\Rightarrow i = i_{g} (G + s)$
$\Rightarrow i = 0.05 (99 + 1) = 5 A$

In case of voltmeter, we have
$V = i_{g} (G + S)$
$V = 0.05 (99 + 101) = 0.05 \times 200 = 10 V$

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