Question

Two 60.0Ω resistors are connected in parallel and this parallel arrangement is then connected in series with a 30.0Ω resistor. The combination is placed across a 120 V potential difference. What is the equivalent resistance and current for the entire circuit?

Answer 1

Answer:-

• R1=R2=60ohm
• R3=30ohm

In parallel

$\boxed{\sf \dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}$

$\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{1}{60}+\dfrac{1}{60}$

$\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{1+1}{60}$

$\\ \sf\longmapsto \dfrac{1}{R}=\dfrac{2}{60}$

$\\ \sf\longmapsto R=\dfrac{60}{2}$

$\\ \sf\longmapsto R=30\Omega$

In series

$\boxed{\sf R=R_1+R_2}$

$\\ \sf\longmapsto R=30+30$

$\\ \sf\longmapsto R=60\Omega$

Now

• Potential difference=V=120V
• Resistance=R=60ohm
• Current=I

According to ohms law

$\boxed{\sf \dfrac{V}{I}=R}$

$\\ \sf\longmapsto I=\dfrac{V}{R}$

$\\ \sf\longmapsto I=\dfrac{120}{60}$

$\\ \sf\longmapsto I=2A$

Answer 2

Answer:

60Ω

Explanation: