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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.