A series combination of two resistors, 9.29 Ω and 2.11 Ω, is connected to a 12.0 V battery. Calculate the equivalent resistance of the circuit. Answer in units of Ω. Calculate the current in the circuit. Answer in units of A.
What is the potential difference across the 9.29 Ω resistor?
Answer in units of V.
What is the potential difference across the 2.11 Ω resistor?
Answer in units of V.
2 answers:
PART A)
Equivalent resistance in series is given as
now we have
PART B)
Here in order to find the current in the circuit we can use ohm's law
here we have
V = 12 Volts
R = 11.40 ohm
PART C)
Now for finding potential difference across 9.29 ohm resistance
PART D)
Similarly for finding potential difference across 2.11 ohm resistance
<h2>Answer:</h2>
<u>Equivalent resistance is 11.4 ohms</u>
<u>The current in the circuit is 1.05 A</u>
<u>The potential difference across the 9.29 Ω resistor is 9.77 v</u>
<u>The potential difference across the 2.11 Ω resistor is 2.21 volts</u>
<h2>Explanation:</h2><h3>Part 1</h3>
The equivalent resistance in series circuit is given by
Re = R1+R2
Re = 9.29 + 2.11
Re = 11.4 Ohms
<h3>Part 2</h3>
According to Ohms law
V = I R
I = V / R
I = 12 / 11.4
I = 1.05 A
<h3>Part 3 </h3>
According to Ohms law
V = I R
V = 1.05 * 9.29
V= 9.77 Volts
<h3>Part 4</h3>
According to Ohms law
V = I R
V = 1.05 * 2.11
V= 2.21 volts
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