Question
Asked Nov 17, 2019
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A student in the laboratory connects a 13 Ω resistor, a 29 Ω resistor, and a 41 Ω resistor in series and then connects the arrangement to a 41-V dc source.

(a) How much current is in the circuit?
___ A

(b) How much power is expended in the circuit?
___ W
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Step 1
(a)Since the resistors are connected in series, the net resistance of the circuit will be equal to the
sum of the individual resistances. The net resistance of the circuit is
R= 13 Ω+ 29 Ω+41 Ω
=83 Q
Here, R is the net resistance of the circuit.
The equation for the current in the circuit is given by
V
I
R
Here, I is the current and V is the potential difference
Substitute the numerical values in the above equation
41 V
I
83 Q
0.494 A
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(a)Since the resistors are connected in series, the net resistance of the circuit will be equal to the sum of the individual resistances. The net resistance of the circuit is R= 13 Ω+ 29 Ω+41 Ω =83 Q Here, R is the net resistance of the circuit. The equation for the current in the circuit is given by V I R Here, I is the current and V is the potential difference Substitute the numerical values in the above equation 41 V I 83 Q 0.494 A

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