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(D) Given,the current through the $20 \ \Omega$ resistor is $i_1 = 0.3 \ A$. The total current measured by the ammeter is $I = i_1 + i_2 + i_3 = 0.9 \

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$30$

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(D) Given,the current through the $20 \ \Omega$ resistor is $i_1 = 0.3 \ A$. The total current measured by the ammeter is $I = i_1 + i_2 + i_3 = 0.9 \ A$.Since the resistors are connected in parallel,the potential difference $V_{AB}$ across them is the same.$V_{AB} = i_1 	imes 20 \ \Omega = 0.3 \ A 	imes 20 \ \Omega = 6 \ V$.Now,calculate the current $i_2$ through the $15 \ \Omega$ resistor:$i_2 = \frac{V_{AB}}{15 \ \Omega} = \frac{6 \ V}{15 \ \Omega} = 0.4 \ A$.Using the total current equation:$i_1 + i_2 + i_3 = 0.9 \ A$$0.3 \ A + 0.4 \ A + i_3 = 0.9 \ A$$0.7 \ A + i_3 = 0.9 \ A$$i_3 = 0.2 \ A$.Finally,calculate $R_1$ using the potential difference $V_{AB}$ and current $i_3$:$R_1 = \frac{V_{AB}}{i_3} = \frac{6 \ V}{0.2 \ A} = 30 \ \Omega$.

In the given circuit,the current flowing through the resistance $20 \ \Omega$ is $0.3 \ A$,while the ammeter reads $0.9 \ A$. The value of $R_1$ is . . . . . . $\Omega$.

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