Question

4. A resistor $R_2$=4.0 $\Omega$ is mounted in parallel with a resistor $R_3$=11.0 $\Omega$, and this combination is connected in series with a resistor $R_1$=10.0 $\Omega$ and a 14 Volts battery. What is the power dissipated in the $R_1$ resistor? On First Attempt = 1 11.7 Watts

Name: 4 a resistor r240 omega is mounted in parallel with a resistor r3110 omega and this combination is connected in series with a resistor r1100 omega and a 14 volts battery what is the power 70819
Uploaded: 2022-10-17T05:11:24-08:00
Duration: 3 min 18 s
Channel: Sri K
Description: 4 a resistor r240 omega is mounted in parallel with a resistor r3110 omega and this combination is connected in series with a resistor r1100 omega and a 14 volts battery what is the power 70819

          4. A resistor $R_2$=4.0 $\Omega$ is mounted in parallel with a resistor $R_3$=11.0 $\Omega$, and this combination is connected in series with a resistor $R_1$=10.0 $\Omega$ and a 14 Volts battery. What is the power dissipated in the $R_1$ resistor?
On First Attempt = 1 11.7 Watts

4 a resistor r240 omega is mounted in parallel with a resistor r3110 omega and this combination is connected in series with a resistor r1100 omega and a 14 volts battery what is the power 70819

Added by Nicholas N.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

10/17/2022

Step 1

The equivalent resistance $R_{23}$ of two resistors $R_2$ and $R_3$ connected in parallel is given by: $\frac{1}{R_{23}} = \frac{1}{R_2} + \frac{1}{R_3}$ $\frac{1}{R_{23}} = \frac{1}{4.0 \Omega} + \frac{1}{11.0 \Omega} = \frac{11.0 + 4.0}{44.0} = Show more…

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4. A resistor $R_2$=4.0 $\Omega$ is mounted in parallel with a resistor $R_3$=11.0 $\Omega$, and this combination is connected in series with a resistor $R_1$=10.0 $\Omega$ and a 14 Volts battery. What is the power dissipated in the $R_1$ resistor? On First Attempt = 1 11.7 Watts