The value of net resistance of the network as | vedclass

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(C) The potential at the left terminal is $V_A = -6 	ext{ V}$ and at the right terminal is $V_B = -8 	ext{ V}$.For the middle branch containing the

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$6 \Omega$

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(C) The potential at the left terminal is $V_A = -6 	ext{ V}$ and at the right terminal is $V_B = -8 	ext{ V}$.For the middle branch containing the $10 \Omega$ resistor and diode $1$,the potential difference across the diode is $V_A - V_B = -6 - (-8) = +2 	ext{ V}$. Since the anode is at a higher potential than the cathode,diode $1$ is in forward bias and acts as a conducting wire.For the bottom branch containing the $5 \Omega$ resistor and diode $2$,the potential difference across the diode is $V_B - V_A = -8 - (-6) = -2 	ext{ V}$. Since the anode is at a lower potential than the cathode,diode $2$ is in reverse bias and acts as an open circuit.The top branch contains a $15 \Omega$ resistor.Thus,the circuit reduces to a parallel combination of the $15 \Omega$ resistor and the $10 \Omega$ resistor.The equivalent resistance $R_{eq}$ is given by:$R_{eq} = \frac{15 	imes 10}{15 + 10} = \frac{150}{25} = 6 \Omega$.

The value of net resistance of the network as shown in the given figure is:

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