Physics Question from JEE Main 2024 · 29 January, Shift 2

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

The figure shows: three resistors in parallel between a left junction and a right junction — $R_1$ in the top branch, $20\,\Omega$ in the middle branch and $15\,\Omega$ in the bottom branch. The supply lead comes in horizontally from the left into the left junction (an arrow marks the incoming current), and the outgoing lead leaves the right junction horizontally and passes through the ammeter $\text{A}$ before going off to the right. The ammeter therefore sits in the main line and carries the total current of all three branches.