In the given circuit, the internal resistance of the cell is zero. If \( i_1 \) and \( i_2 \) are the readings of the ammeter when the key (K) is opened and closed respectively, then \( i_1 : i_2 = \)
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In circuits with resistors in parallel and series, always reduce the network step-by-step to equivalent resistance before applying Ohm’s law.
Step 1: When the key \(K\) is open, only one 40\(\Omega\) resistor is in the circuit. Hence, the total resistance is \( R = 40\Omega \).
\[
i_1 = \frac{V}{R} = \frac{12}{40} = 0.3\text{ A}
\]
Step 2: When the key \(K\) is closed, the two 40\(\Omega\) resistors are in parallel:
\[
R_{\text{parallel}} = \frac{40 \times 40}{40 + 40} = 20\Omega
\]
Now,
\[
i_2 = \frac{12}{20} = 0.6\text{ A}
\]
Step 3: Therefore, the ratio is:
\[
i_1 : i_2 = \frac{0.3}{0.6} = \frac{1}{2}
\]
