I=VcdR3≈0.17A,Vab=V−Vcd=2V,I=\frac{V_{cd}}{R_3}\approx0.17A, V_{ab}=V-V_{cd}=2V,I=R3Vcd≈0.17A,Vab=V−Vcd=2V, hence I1=Vab/R1=0.1AI_1=V_{ab}/R_1=0.1AI1=Vab/R1=0.1A hence
I2=I−I1=0.07AI_2=I-I_1=0.07AI2=I−I1=0.07A hense R2=V1/I2≈29ohmsR_2=V_1/I_2\approx29ohmsR2=V1/I2≈29ohms
where R1=20ohms,R3=24ohms,Vcd=4V,V=6VR_1=20ohms,R_3=24ohms, V_{cd}=4V,V=6VR1=20ohms,R3=24ohms,Vcd=4V,V=6V
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