Question

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.

Answer 1

PART A)

Equivalent resistance in series is given as

$R = R_1 + R_2$

now we have

$R = 9.29 + 2.11$

$R = 11.40 ohm$

PART B)

Here in order to find the current in the circuit we can use ohm's law

$V = iR$

here we have

V = 12 Volts

R = 11.40 ohm

$12 = (i)(11.40 ohm)$

$i = 1.05 A$

PART C)

Now for finding potential difference across 9.29 ohm resistance

$V = iR$

$V = (1.05 A)(9.29 ohm)$

$V = 9.78 Volts$

PART D)

Similarly for finding potential difference across 2.11 ohm resistance

$V = iR$

$V = (1.05 A)(2.11 ohm)$

$V = 2.22 Volts$

Answer 2

Answer:

Equivalent resistance is 11.4 ohms

The current in the circuit is 1.05 A

The potential difference across the 9.29 Ω resistor is 9.77 v

The potential difference across the 2.11 Ω resistor is 2.21 volts

Explanation:

Part 1

The equivalent resistance in series circuit is given by

Re = R1+R2

Re = 9.29 + 2.11

Re = 11.4 Ohms

Part 2

According to Ohms law

V = I R

I = V / R

I = 12 / 11.4

I = 1.05 A

Part 3

According to Ohms law

V = I R

V = 1.05 * 9.29

V= 9.77 Volts

Part 4

According to Ohms law

V = I R

V = 1.05 * 2.11

V= 2.21 volts