Interpret the circuit information

Since the actual circuit image is not shown here, we use the given data in the standard way:

• Current through the $20\,\Omega$ resistor is $0.3\,\text{A}$.
• Ammeter reads total current $0.9\,\text{A}$.

This implies the ammeter is measuring the total current entering a parallel combination containing:

• one branch with $20\,\Omega$
• another branch with $R_1$

Find current through $R_1$

Total current is the sum of branch currents: $I=I_{20}+I_{R_1}$ $0.9=0.3+I_{R_1}$ $I_{R_1}=0.6\,\text{A}$

Find voltage across the parallel branches

In parallel, the voltage across each branch is the same.

For the $20\,\Omega$ branch: $V=IR=(0.3)(20)=6\,\text{V}$

So voltage across $R_1$ is also $6\,\text{V}$.

Calculate $R_1$

Using Ohm’s law: $R_1=\frac{V}{I_{R_1}}=\frac{6}{0.6}=10\,\Omega$

Final answer

$\boxed{R_1=10\,\Omega}$

Compare with stored correct answer

Stored correct answer is $30\,\Omega$, but the given numerical data lead to: $R_1=10\,\Omega$

So I disagree with the stored answer. The stored answer may correspond to a different circuit arrangement than what is implied by the given currents, but with the stated values, $10\,\Omega$ is the consistent result.