If each diode in figure has a forward bias resistance of $25 \Omega$ and infinite resistance in reve

Question

If each diode in figure has a forward bias resistance of $25 \Omega$ and infinite resistance in reverse bias, the values of the current

Professor PC Thomas Classes · Answer

(d) : Let $R$ be the effective resistance of the circuit, then $R=\left(R_{A B} \| R_{E F}ight)+25$ \ $R_{A B}=125+25=150 \Omega ; R_{E F}=125+25=150 \Omega$\ $	herefore \quad R=25+\frac{150}{2}=100 \Omega$\ Since diode in the branch $C D$ is reverse biased, $I_3=0$. Current, $I_1=\frac{5}{100}=0.05 \mathrm{~A}$ According to Kirchhoff's current rule,  \ $I_1=I_2+I_3+I_4$ or $I_2+I_4=I_1=0.05 \mathrm{~A}$\ $\because \quad R_{A B}=R_{E F}$, so, $I_4=I_2$\ $2 I_4=2 I_2=0.05 ; I_4=I_2=\frac{0.05}{2}=0.025 \mathrm{~A}$