Question

The emf ε and the internal resistance r of the battery, shown in the figure, are 4.3 V and 1.0 Ω respectively. The external resistance R  is 50 Ω. The resistances of the ammeter and voltmeter are 2.0 Ω and 200 Ω respectively. (a) Find the readings of the two meters. (b) The switch is thrown to the other side. What will be the readings of the two meters now?

Answer 1

(a)

The 50 Ω and 200 Ω resistances are in parallel. Their equivalent resistance,

\[R_{eqv} = \frac{50 \times 200}{50 + 200} = 40 \Omega\]

This equivalent resistance and the 2 Ω and 1 Ω resistances are connected in series. The effective resistance of the circuit,

\[R_{eff} = \left( 40 + 2 + 1 \right) = 43 \Omega\]

In this case, the ammeter will read the total current of the circuit. The current through the ammeter,

\[i = \frac{4 . 3}{43} = 0 . 1 A\]

This current will be distributed in the inverse ratio of resistance between the resistances 50 Ω and 200 Ω. The current through the voltmeter,

\[i' = \frac{50}{\left( 50 + 200 \right)}i\]

\[ \Rightarrow i' = 0 . 02 A\]

Reading of the voltmeter = 0.02 × 200 = 4 V

(b)

The two branches AB and CD are in parallel. Their equivalent resistance,

\[R_{eqv} = \frac{52 \times 200}{52 + 200} = 41 . 27 A\]

This equivalent resistance is in series with the 1 Ω resistance. The effective resistance of the circuit,

\[R_{eff} = \left( 41 . 27 + 1 \right) \Omega = 42 . 27 \Omega\]

The total current through the circuit,

\[i = \frac{4 . 3}{42 . 27} = 0 . 1 A\]

In this case, the ammeter will read the current flowing through the 50 Ω resistance, which is i₁, as shown. The currents in the two parallel branches will distribute in the inverse ratio of the resistances.

\[\therefore i_1 = \left( \frac{200}{200 + 52} \right)i\]

\[ \Rightarrow i_i = 0 . 08 A\]

The current through the voltmeter = i - i₁ = 0.02 A

The reading of the voltmeter = 0.02 × 200 = 4 V