ECE 1100: Introduction to Electrical and Computer Engineering Notes 22 Voltage and Current Divider Rules Spring 2008 David R. Jackson Professor, ECE Dept.

Voltage Divider Rule + - v s ( t ) R1R1 R2R2 + - v 2 ( t ) Find v 2 ( t ) Note: there is an open circuit at the output terminals.

Voltage Divider Rule + - v s ( t ) R1R1 R2R2 + - v 2 ( t ) i ( t )

Voltage Divider Rule Rule: The multiplying factor that gives the voltage across a resistor is the resistance of the resistor divided by sum of the resistances. + - v s ( t ) R1R1 R2R2 + - v 2 ( t )

Example a)When measured with an ideal voltmeter (infinite internal resistance), b)When measured with a digital multimeter (DMM) that has a 10 [ M  ] resistance [V] R1R1 R2R2 + - V2V2 R 1 = 100 [  ] R 2 = 200 [  ] Find the voltage V 2 :

R 1 = 100 [  ] R 2 = 200 [  ] (a) [V] R1R1 R2R2 + - V2V2 Example (cont.)

R 1 = 100 [  ] R 2 = 200 [  ] (b) [V] R1R1 R2R2 + - V2V2 R DMM R DMM = 10 [M  ] Note that R 2 and R DMM are in parallel, so we can combine them. Example (cont.)

R 1 = 100 [  ] R 2 = 200 [  ] (b) R DMM = 10 [M  ] [V] R1R1 R2R2 + - V2V2 R DMM Example (cont.)

R 1 = 100 [  ] R 2 = 200 [  ] (b) R eq = [  ] [V] R1R1 R eq + - V2V2 Example (cont.)

R 1 = 100 [M  ] R 2 = 200 [M  ] Example (cont.) Try the same example again using: [V] R1R1 R2R2 + - V2V2 R DMM Ideal DMM: Actual DMM: There is a huge amount of loading by the DMM!

Current Divider Rule v s ( t ) + - R1R1 R2R2 i ( t ) i1 (t)i1 (t) i 2 ( t ) Find i 2 ( t )

Current Divider Rule v s ( t ) + - R1R1 R2R2 i ( t ) i1 (t)i1 (t) i 2 ( t )

Current Divider Rule Rule: The multiplying factor that gives the current through a resistor is the opposite resistance divided by the sum of the two resistances. v s ( t ) + - R1R1 R2R2 i ( t ) i1 (t)i1 (t) i 2 ( t )

Example Find the RMS voltage across the two appliances and the current coming out of the outlet. Also find the current through R L2 and the average power dissipated by R L2. I2I2 + - V s =120 [V] (RMS) R s = 3 [  ] RL1RL1 + - V2V2 RL2RL2 R L1 = 144 [  ] R L2 = 14.4 [  ] R L1 = 100 [ W ] light bulb R L2 = 1000 [ W ] hair dryer R s = resistance of house wiring outlet

Example (cont.) R L1 = 144 [  ] R L2 = 14.4 [  ] + - V s =120 [V] (RMS) R s = 3 [  ] RL1RL1 + - V2V2 RL2RL2 I2I2

+ - V s =120 [V] (RMS) R s = 3 [  ] R L eq = [  ] + - V2V2 Example (cont.)

+ - V s =120 [V] (RMS) R s = 3 [  ] R L eq = [  ] + - V2V2 IsIs (This is the current (RMS) coming out of the outlet.)

R L1 = 144 [  ] R L2 = 14.4 [  ] + - V s =120 [V] R s = 3 [  ] RL1RL1 + - V2V2 RL2RL2 I s = [A] I2I2 Example (cont.)

Note: If there was 120 [ V ] (RMS) across the hair dryer, we would have