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(D) The resistors $R_1$, $20 \, \Omega$, and $15 \, \Omega$ are connected in parallel. Therefore, the voltage $V$ across each resistor is the same.The

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$60$

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(D) The resistors $R_1$, $20 \, \Omega$, and $15 \, \Omega$ are connected in parallel. Therefore, the voltage $V$ across each resistor is the same.The voltage across the $20 \, \Omega$ resistor is given by $V = I_{20} 	imes R_{20} = 0.3 \, A 	imes 20 \, \Omega = 6 \, V$.Since the resistors are in parallel, the voltage across the $15 \, \Omega$ resistor is also $6 \, V$. The current through the $15 \, \Omega$ resistor is $I_{15} = \frac{V}{R_{15}} = \frac{6 \, V}{15 \, \Omega} = 0.4 \, A$.The total current measured by the ammeter is $I_{total} = 0.8 \, A$. According to Kirchhoff's Current Law, $I_{total} = I_{R1} + I_{20} + I_{15}$.Substituting the known values: $0.8 \, A = I_{R1} + 0.3 \, A + 0.4 \, A$.$I_{R1} = 0.8 \, A - 0.7 \, A = 0.1 \, A$.Since the voltage across $R_1$ is also $6 \, V$, the value of $R_1$ is $R_1 = \frac{V}{I_{R1}} = \frac{6 \, V}{0.1 \, A} = 60 \, \Omega$.

In the given circuit, the current flowing through the resistance $20 \, \Omega$ is $0.3 \, A$, while the ammeter reads $0.8 \, A$. What is the value of $R_1$ in $\Omega$?

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