Two resistors, of R1 = 3.49 ? and R2 = 6.93 ?, are connected in series to a batt

Question

Two resistors, of R1 = 3.49 ? and R2 = 6.93 ?, are connected in series to a battery with an EMF of 24.0 V and negligible internal resistance. Find the current I1 through the first resistor and the potential difference ?V2 between the ends of the second resistor.

Please provide BOTH answers! Thank you!!

Two resistors, of R1 = 3.49 and R2-6.93 , are connected in series to a battery with an EMF of 24.0 V and negligible internal resistance. Find the current h through the first resistor and the potential difference ½ between the ends of the second resistor R1-3.49 Number 24.0 V Number R,=693

Explanation / Answer

Rnet = R1 + R2 = 3.49 + 6.93 = 10.42 ohm

I = V / Rnet = 24 / 10.42 = 2.3 A

I is same in series so I = I1 = 2.3 A

V across R2 = I R2 = 2.3 x 6.93 = 15.9 V

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---