Four equal resistors,each of resistance 10 ,O | vedclass

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(B) The circuit consists of two parallel branches connected between points $A$ and $B$. Each branch contains two $10 \,\Omega$ resistors in series. $1

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$10$

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(B) The circuit consists of two parallel branches connected between points $A$ and $B$. Each branch contains two $10 \,\Omega$ resistors in series. $1$. The upper branch has two resistors of $10 \,\Omega$ in series,so its resistance is $R_1 = 10 \,\Omega + 10 \,\Omega = 20 \,\Omega$.$2$. The lower branch also has two resistors of $10 \,\Omega$ in series,so its resistance is $R_2 = 10 \,\Omega + 10 \,\Omega = 20 \,\Omega$.$3$. The diode is connected between the midpoints of these two branches. Since the circuit is symmetric,the potential at both midpoints is the same,meaning there is no potential difference across the diode. Therefore,the diode is in a state of equilibrium and no current flows through it.$4$. The equivalent resistance $R_{eq}$ between $A$ and $B$ is the parallel combination of $R_1$ and $R_2$:$R_{eq} = \frac{R_1 	imes R_2}{R_1 + R_2} = \frac{20 	imes 20}{20 + 20} = \frac{400}{40} = 10 \,\Omega$.

Four equal resistors,each of resistance $10 \,\Omega$,are connected as shown in the circuit diagram. The equivalent resistance between $A$ and $B$ is ............. $\Omega$.

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