KIRCHHOFF’S LAWS & COMBINATIONS OF RESISTORS

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Presentation transcript:

KIRCHHOFF’S LAWS & COMBINATIONS OF RESISTORS RESISTORS IN SERIES RESISTORS IN PARALLEL THE POTENTIAL DIVIDER © John Parkinson

THE ALGEBRAIC SUM OF THE CURRENTS KIRCHHOFF’S FIRST LAW I = 0 THE ALGEBRAIC SUM OF THE CURRENTS AT A JUNCTION IS ZERO I3 I4 I1 I2 I5 I1 + I2 = I3 + I4 + I5 © John Parkinson

PUTTING IT SIMPLY EQUALS CURRENT OUT CURRENT IN EQUALS CURRENT OUT 0.5 A Example ? A 0.3 A 0.5 A - 0.3 A = 0.2 A © John Parkinson

KIRCHHOFF’S SECOND LAW AROUND A CLOSED CIRCUIT LOOP, THE ALGEBRAIC SUM OF THE e.m.f.s IS EQUAL TO THE ALGEBRAIC SUM OF THE p.d.s  R1 R2 R3 R4 I IR1 IR2 (I-I4)R3 I I-I4 I4 Then  = IR1 + IR2 + (I-I4) R3 For R3 and R4 loop , 0 = I4R4 - (I-I4) with no e.m.f.’s in this loop © John Parkinson

Example A battery charger has an e.m.f. of 14 V and an internal resistance of 1.6  . It is used to charge a 12 V battery, which itself has an internal resistance of 0.3 , through a 1  resistor. Find the charging current. Charger  = 14V-12V = 2V 0.1  1.6  Sum of p.d.s =  IR Ix0.3 + Ix0.1 + Ix1.6 = Ix2 12 V 14 V So 2V = Ix2 0.3  Charger I = 1A I © John Parkinson

By Kirchhoff’s SECOND law RESISTORS IN SERIES e.m.f.=  R1 R2 R3 I  = IR1 + IR2 + IR3 = I(R1+R2+R3) By Kirchhoff’s SECOND law If we apply Ohm’s Law to the whole circuit, we have  = IR, if R is the total resistance So IR = I(R1+R2+R3) R = R1+R2+R3 Dividing each side by I © John Parkinson

By Kirchoff’s FIRST law RESISTORS IN PARALLEL e.m.f. =  By Kirchoff’s FIRST law  I = I1 + I2 + I3 I I1 R1 We now apply Ohm’s Law to each component and to the whole circuit, letting R = the total resistance I2 R2 I3 R3 DIVIDING EACH SIDE BY V © John Parkinson

Example Find the current in each resistor Source = 20V What is the combined resistance of B and C? Source = 20V IB B=12 So the total resistance in the circuit, including A, is 4 + 6  = 10 A=6 8V C=6 What is the current taken from the battery & the current in A? IC 20V  10  = 2A The p.d. across B & C is given by V = IR = 2A X 4  = 8V Hence find IB and IC IB = V  R = 8  12 = 2/3 Amps IC = V  R = 8  6 = 4/3 Amps © John Parkinson

in the same ratio a the resistors THE POTENTIAL DIVIDER A If NO current is drawn through the output, then by Ohm’s Law I R1 Vin B R2 Vout and Vout = IxR2 C so The voltage is divided in the same ratio a the resistors © John Parkinson

Example 150V A Vout B So RAB = 5 Ohms Vout = 75 V WHY? (i) Find Vout k A 10 k Vout (ii) Find Vout if another 10 k resistor is connected across AB B So RAB = 5 Ohms Vout = 75 V WHY? © John Parkinson