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We can solve this circuit by using Kirchoff's current and voltage laws Kirchoff's current law states that the current coming into a junction should be equal to the current flowing out of the junction. Kirchoff's voltage law states that the sum of the emf's around any closed loop should be zero In the shown circuit, we can write the following junction formula I1=I2+I3 Let this be equation no 1For kirchoffs voltage law, we will consider 3 closed loops in the circuit. The first loop on the left side, the second loop on the right side and the third loop along the perimeter of the circuit. For using sign conventions, we will use the following points A drop in voltage will give a negative sign and an increase in voltage will give a positive sign. As we move from the positive to negative terminal of a battery, the potential drops and hence a negative sign will be accompanied with the voltage of this battery and vice versa. If we travel along the loop in the direction of the current, and there is a resistor, then the voltage across the resistor will have a negative sign convention since there will be a voltage drop across the resistor. And vice versa. Consider the first loop. +10-2I2+5-4(I2+I3)=0+10-2I2+5-4I1=04I1+2I2=15Let this be equation 2 For the second loop -5I3-3-5+2I2=02I2-5I3=8Let this be equation 3 From equation 1, I1=I2+I3I2=I1-I3Using this value of I2 in equations 2 and 3,4I1+2I1-2I3=156I1-2I3=152I1-2I3-5I3=82I1-7I3=8Solving these two equations simultaneously,21I3-2I3=-24+1519I3=-9I3=-0.473 A Negative sign indicates that the direction of I3 is opposite of that shown in the diagram From I3, we can find the remaining two currents as 2I1-7I3=82I1-7(-0.473)=82I1=8-3.311I1=4.6892I1=2.3445 AI2=I1-I3I2=2.3445-(-0.473)I2=2.8175 A As we have all the three values of currents, we can find the voltages across the 3 resistors as VA=I1*4VA=2.3445*4VA=9.378 VVB=2*I2VB=2*2.8175VB=5.635 VVC=5*I3VC=5*0.473VC=2.365 V In order to find the total power across the circuit, we will find the individual powers dissipated along the 3 resistors P1=I1VAP1=2.3445*9.378P1=21.986721 WattsP2=I3VCP2=0.473*2.365P2=1.118645 WattsP3=I2VBP3=2.8175*5.635P3=15.8766125 Watts Now, the total power dissipated in the circuit isP=P1+P2+P3P=21.986721+1.118645+15.8766125P=38.98 WAnswers: I1 = 2.3445 A I2 = 2.8175 A I3 = 0.473 A VA = 9.378 V VB = 5.635 V VC = 2.365 V Ptotal = 38.98 W

Answered: し。 Tov しょ 5V A V = TOT

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Advanced Physics

し。 Tov しょ 5V A V = TOT

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