Kirchoff's Laws
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Example #1 Problem: Find the currents through all the resistors in the circuit below: ![]() DATA: Vb = 12 V, R1 = 10 W, R2 = 15 W, R3 = 20 W Solution: Summing the voltages around the left and right loops gives the following two equations: where i3 has been replaced by i1 - i2. Multiplying Eq. (1) by R3, multiplying Eq. (2) by R1, then adding the equations yields: which rearranged yields Once i2 is known, Eq. (1) can be used to get i1, and i3 can be found as the difference i1 - i2.i2 = 0.554 amps, i1= .369 amps, i3 = -.185 amps Example #2 Problem: Find the charges on all the capacitors in the circuit below: ![]() DATA: Vb = 12 V, C1 = 10 mF, C2 = 15 mF, C3 = 20 mF Solution: Summing the voltages around the left and right loops gives the following two equations where Q3 has been replaced by Q1 - Q2. Dividing Eq. (1) by C3, dividiing Eq. (2) by C1, then adding the equations yields: ![]() which rearranged yields ![]() Once Q2 is known, Eq. (1) can be used to get Q1, and Q3 can be found as the difference Q1 - Q2. Q2 = 120.0 mC, Q1= 40.0 mC, Q3 = -80 mC Example #3 The circuit below has been in position a for a long time. At time t = 0 the switch is thrown to position b. DATA: Vb = 12 V, C = 10 mF, R = 20 W ![]() a.) What is the curnent through the resistor just BEFORE the switch is thrown? I = 0 b.) What is the current through the resistor just AFTER the switch is thrown? Solution: I = V/R I = 0.6 amps c.) What is the charge across the capacitor just BEFORE the switch is thrown? Solution: Q = CV Q = 120 mC d.) What is the charge on the capacitor just AFTER the switch is thrown? Solution: Charge does not change instantaneously. Q = 120 mC e.) What is the charge on the capacitor at at time t = 0.3 msec after the switch is thrown? Solution: Q = Q0exp(-t/t) , where t = RC = 0.2 msec Q = 26.8 mC Example #4 Considering the same circuit, only with the switch thrown from b to a at time t = 0 after having been in position b for a long time. DATA: Vb = 12 V, C = 10 mF, R = 20 W ![]() a.) What is the curnent through the resistor just BEFORE the switch is thrown? I = 0 b.) What is the current through the resistor just AFTER the switch is thrown? Solution: I = V/R I = 0.6 amps c.) What is the charge across the capacitor just BEFORE the switch is thrown? Solution: Q = CV Q = 0 d.) What is the charge on the capacitor just AFTER the switch is thrown? Solution: Charge does not change instantaneously. Q = 0 e.) What is the charge on the capacitor at at time t = 0.3 msec after the switch is thrown? Solution: Q = Q0(1.0 - exp(-t/t)) , where t = RC = 0.2 msec Q = 93.2 mC
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Once i2 is known, Eq. (1) can be used to get i1, and i3 can be found as the difference i1 - i2.













