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Solutions to parallel circuit problems involving finding total resistance, current passing through each resistor, and missing readings. It includes calculations for various resistor combinations and voltage inputs.
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Parallel Circuits Solutions to p 126b,c
Rt = [R 1 -1^ + R 2 -1^ + R 3 -1^ ]-1^ = [21-1^ + 21 -1^ + 21 -1^ ]-1^ = 7Ω.
b. 5 Ω
Rt = [R 1 -1^ + R 2 -1^ + R 3 -1^ ]-1^ = [5-1^ + 10 -1^ + 20 -1^ ]-1^ = 2.9 Ω.
c. 100 Ω
Rt = [R 1 -1^ + R 2 -1^ + R 3 -1^ ]-1^ = [100-1^ + 50 -1^ + 100 -1^ ]-1^ = 25 Ω.
a. 12 Ω
Since voltage is constant in parallel, I 1 = V/R 1 = 120/12 = 10A I 2 = V/R 2 = 120/12 = 10A I 3 = V/R 3 = 120/12 = 10A
What is the current intensity ( I ) flowing through the ammeter?
I = 20/40 = 0.50 A
R (^1)
R (^2)
R (^3)
1 3
What is the value of resistor R 3?
Since voltage is always constant in parallel, V = 3(4) or (6)(2) = 12V. R 3 = 12V/6A = 2 Ω.
Switch Light Bulb
S 1 S 2 L 1 L 2
open open out out
closed open bright out
Which of the following circuit diagrams illustrates the results shown in the table above?
A) C)
