Use KVL KCL and an appropriate circuit solving technique...

Use KVL & KCL and an appropriate circuit solving technique to solve for the currents and voltages in each element of the following circuit: Compute the power dissipated by each resistor Compute the power supplied by each battery a b c d $R_3 = 20.0 Omega$ $R_1 = 10.0 Omega$ $V_2 = 24.0 V$ $V_1 = 12.0 V$ $R_2 = 30.0 Omega$

Transcript

00:01 Hello students so in this question we have to calculate the current and voltage in each elements of the circuit and the circuit is shown below a b here we have a resistance of 20 oms here we have a battery of 24 volts at this point is c here we have another resistance of resistance 30 oms here we have point b here we have another resistance of 10 oms we have given another battery of 12 volt voltage.

00:46 Okay, so we have to calculate the net current and voltage to across all of these elements.

00:52 Okay, so let's assume that the current is flowing like this and the current value has a magnitude of i.

00:58 So the direction of current can be represented like this.

01:02 So we can use kirchof voltage low like this.

01:08 The potential difference from we'll choose a loop ab d c a so potential difference from a b c sorry a b d c a b now this will be equal to 0 delta v okay so we'll write from a to b from a to b there is no resistance or anything like this just simple via so potential difference will not drop okay so from b to d we have a resistance and the current flowing is i so we can write minus i r that is the potential drop across this resistor okay so r has a resistance of 10 oms so we can write minus 10 i now plus here we have the positive terminal of the battery and here we have negative terminal of the battery so this potential drop will be plus 12 here again we have current i and a resistance so across this resistance potential drop will be equal to current times its resistance that is minus 30 i plus here we have another battery here it's is its negative terminal and here we have positive terminal so the potential drop across this battery will be minus 24 so minus 24 volt now again we have a resistor and a current i is flowing through it so the potential drop across it will be minus 20 i now this will be equal to from here we can write that 20 i plus 30 i 50 i 50 i 50 i plus 10 i 60 i so 60 i plus 24 now this will be equal to 12 so we can write that 60 i is equals to 12 minus 24 that is minus 12 from here we can write the value of current as minus 1 divided by 5 mpm the minus here represents the direction of current should be opposite to this shown so the actual direction of the current will be like this and this will have a magnitude of minus i that means plus 1 .1 divided by 5mps so this is the value of current and we have to find the voltage across each element so the voltage across these batteries is already given 24 and 12 volt we just have to find the voltage across these resistors okay so let's say the voltage across 20 -oom resistor v20 oom now this will be equal to the current flowing through this resistor times the resistance of this resistor so we can say the voltage across this will be 1 divided by 5 times its resistance that is 20 om so this value will come out to be equal to 4 volts similarly voltage across 30 -hom resistor will be equal to the current that is 1 by 5 ampers times its resistance that is 30 oms so this will be equal to 6 volts...

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