11. The current in the 9.00 Volt battery is equal to 12. The potential differenc
ID: 1448504 • Letter: 1
Question
11. The current in the 9.00 Volt battery is equal to
12. The potential difference across the terminals of battery #1 from point a to point b is equal to
13. The potential difference across the terminals of battery #2 from point c to point d is equal to
Please show ALL work! Thank you.
The circuit below contains two real batteries, each having significant internal resistance. Battery #1 has an EMF of 9.00 V and an internal resistance of 1.00 ohm. Battery #2 has an EMF of 3.00 V and an internal resistance of 2.00 ohms. These two batteries are connected in series with two resistors R1 = 5.00 ohms and R2-4.00 ohms. These two batteries are connected in series with two resistors R-5.00 ohms and R,-4.00 ohms. Ri (5.00 2) 9.00 V) (3.00 V) T1 (1 .0012) 1 2 172 (4.00) (2)Explanation / Answer
11)
let I is the current through the ckt.
Apply KVL in the loop
E1 - I*R1 - E2 - I*r2 - I*R2 - I*r1 = 0
I*(R1 + R2 + r1 + r2) = (E1 - E2)
I = (E1-E2)/(R1 + R2 + r1 + r2)
= (9 - 3)/(5 + 4 + 1 + 2)
= 0.5 A
12) V1_terminal = E1 - I*r1
= 9 - 0.5*1
= 8.5 volts
13) V2_terminal = E2 - I*r2
= 8 - 0.5*2
= 7 volts
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.