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Question 3. Find the total resistance of the following circuit.
Formulas:
R= V/I
If the # of resistors connected in series is n, the equiv. resistor is given by
Rs = R1 + R2 +... + Rn
likewise with resistors connected in parallel,
1/Rp = 1/R1 + 1/R2 +... + 1/Rn
Given:
V0 = 120 V
R1 = 20 Ω
R2 = 22 Ω
R3 = 30 Ω
R4 = 27 Ω
R5 = 40 Ω
R6 = 20 Ω
R7 = 20 Ω
My attempt:
Applying Ohm’s Law to the entire circuit:
I0 = V0 / RT , where RT is the total resistance in the circuit
But since R3 and R4 are connected in series,
RS = R3 + R4
= 30 Ω + 27 Ω
= 57 Ω
Then RT1 and R7 are connected in parallel:
1/RP = 1/RS + 1/R7
= 1/57 + 1/20
= 20/1140 + 57/1140
= 77/1140 Ω
RP = 14.8 Ω
= 15 Ω
Solve five resistors R1, R2, RP, R5 and R6. These resistors are connected in series
RT = R1 + R2 + R3 + R4 + R5 + R6
= 20 Ω + 22 Ω + 15 Ω + 40 Ω + 20 Ω
= 117 Ω
Therefore, the total resistance of the circuit is 117 Ω.
I'm not sure I took the right approach. Any help would be appreciated. Thanks.
Formulas:
R= V/I
If the # of resistors connected in series is n, the equiv. resistor is given by
Rs = R1 + R2 +... + Rn
likewise with resistors connected in parallel,
1/Rp = 1/R1 + 1/R2 +... + 1/Rn
Given:
V0 = 120 V
R1 = 20 Ω
R2 = 22 Ω
R3 = 30 Ω
R4 = 27 Ω
R5 = 40 Ω
R6 = 20 Ω
R7 = 20 Ω
My attempt:
Applying Ohm’s Law to the entire circuit:
I0 = V0 / RT , where RT is the total resistance in the circuit
But since R3 and R4 are connected in series,
RS = R3 + R4
= 30 Ω + 27 Ω
= 57 Ω
Then RT1 and R7 are connected in parallel:
1/RP = 1/RS + 1/R7
= 1/57 + 1/20
= 20/1140 + 57/1140
= 77/1140 Ω
RP = 14.8 Ω
= 15 Ω
Solve five resistors R1, R2, RP, R5 and R6. These resistors are connected in series
RT = R1 + R2 + R3 + R4 + R5 + R6
= 20 Ω + 22 Ω + 15 Ω + 40 Ω + 20 Ω
= 117 Ω
Therefore, the total resistance of the circuit is 117 Ω.
I'm not sure I took the right approach. Any help would be appreciated. Thanks.
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