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(A) For a Wheatstone bridge to be balanced,the ratio of resistances in the two arms must be equal,i.e.,$\frac{R_1}{R_2} = \frac{R_3}{R_4}$.In the give

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(A) For a Wheatstone bridge to be balanced,the ratio of resistances in the two arms must be equal,i.e.,$\frac{R_1}{R_2} = \frac{R_3}{R_4}$.In the given circuit,the left arm has resistances $10 \Omega$ and $15 \Omega$,so the ratio is $\frac{10}{15} = \frac{2}{3}$.For the right arm,we need the effective resistance $R_{eff}$ such that $\frac{10}{R_{eff}} = \frac{2}{3}$,which implies $R_{eff} = 15 \Omega$.The right arm consists of a $5 \Omega$ resistor in series with a diode $D$. If the diode is forward-biased and has a dynamic resistance $R_D = 10 \Omega$,the total resistance becomes $R_{eff} = 5 \Omega + 10 \Omega = 15 \Omega$.Thus,the bridge is balanced when the diode is in series with the $5 \Omega$ resistor and is forward-biased. Looking at the options,the circuit where the diode is in series with the $5 \Omega$ resistor and oriented to be forward-biased allows for this balance.

Choose the correct circuit which can achieve the bridge balance.

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