In the circuit shown the variable resistance is so adjusted that the ammeter reading is same in both the position $1$ and $2$ of the key. The reading of ammeter is $2A$. If $E = 20V$, then $x$ is :- ................... $\Omega$
Medium
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$\mathrm{E}-\mathrm{Ix}=0 \Rightarrow \mathrm{x}=\frac{\mathrm{E}}{\mathrm{I}}=\frac{20}{2}=10 \,\Omega$
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