Asked by Vicki

In a circuit, a 10 Ω resistor is connected in series to a parallel group containing a 60 Ω resistor and a 5 Ω resistor. What is the total resistance in this circuit?

A formula and explanation would be great!
Answered by DrBob222
The parallel group has an equivalent resistance of
1/R = 1/R1 + 1/R2
1/R = 1/60 + 1/5
R = ? which I will call Rp

Then the total for the circuit is Rtotal = 10 + Rp = ?
Answered by Henry
Another Method
R1 = 10 Ohms
R2 = 60 Ohms
R3 = 5 Ohms

Rt = R1 + (R2*R3)/(R2+R3)
Answered by Vicki
I think I am just dumb. Lol, I came up with 204.7 which is too big and not an option. What am I doing wrong. I know I'm doing something wrong... :(
Answered by Vicki
Never mind! Answer is 830Ω

Thank you!
Answered by Henry
Rt = 10 * (60*5)/(60+5) = 10 + 4.62 = 14.62 Ohms = Total resistance.

Note: Based on the values given, the total resistance will be less than 15
ohms.
Answered by Henry
Correction:
Rt = 10 + (60*5)/(60+5) = 10 + 4.62 =
14.62 Ohms.
Answered by Vicki
The answer was 14.615Ω lol, the 830 was for another problem I was having difficulty with! I got both right though, so thank you! I just got mixed up here.
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