Assume that R1 is connected in series with R2 and they form one branch, and R3 is in series with R4, which form another branch.
The branches are in parallel. Then the total resistance will be
R=R1+R2+R3+R4(R1+R2)(R3+R4).According to Ohm's law, the supply voltage is total resistance times total current:
V=IR.V=IR1+R2+R3+R4(R1+R2)(R3+R4),R2=429 V.Calculate the voltage drops across each resistor. As we know, common voltages across resistors [1 in series with 2] and [3 in series with 4] are equal:
V=I1(R1+R2)V=I2(R3+R4),I1=0.0532 A,I2=0.264 A. The voltage drops:
VR1=I1R1=1.17 V,VR2=IR2=22.8 V,VR3=IR3=11.4 V,VR4=IR4=12.7 V,Vt=V=VR1+VR2=VR3+VR4=24 V.
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