Question #125043
Circuit battery is 24V and R1 is 22 ohms, R2 is ? , R3 is 43 ohms and R4 is 48 ohms. The total supply current is 317ma. Find Vt, Vr1, Vr2, Vr3, Vr4
1
Expert's answer
2020-07-06T15:39:54-0400

Assume that R1R_1 is connected in series with R2R_2 and they form one branch, and R3R_3 is in series with R4R_4, which form another branch.



The branches are in parallel. Then the total resistance will be


R=(R1+R2)(R3+R4)R1+R2+R3+R4.R=\frac{(R_1+R_2)(R_3+R_4)}{R_1+R_2+R_3+R_4}.

According to Ohm's law, the supply voltage is total resistance times total current:


V=IR.V=I(R1+R2)(R3+R4)R1+R2+R3+R4,R2=429 V.V=IR.\\ V=I\frac{(R_1+R_2)(R_3+R_4)}{R_1+R_2+R_3+R_4},\\ R_2=429\text{ 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.V=I_1(R_1+R_2)\\ V=I_2(R_3+R_4),\\ I_1=0.0532\text{ A},\\ I_2=0.264\text{ 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.V_{R_1}=I_1R_1=1.17\text{ V},\\ V_{R_2}=IR_2=22.8\text{ V},\\ V_{R_3}=IR_3=11.4\text{ V},\\ V_{R_4}=IR_4=12.7\text{ V},\\ V_t=V=V_{R_1}+V_{R_2}=V_{R_3}+V_{R_4}=24\text{ V}.

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