A circuit is constructed with five resistors and one real battery as shown above
ID: 1415631 • Letter: A
Question
A circuit is constructed with five resistors and one real battery as shown above right. We model. The real battery as an ideal emf V-12 V in series with an internal resistance r as shown above left. The values for the resistors are: R1-R3 280, R4-R5-90 and R2-810. The measured voltage across the terminals of the batery is Vbattery-11.58 V. I, R RI I R R4 R4 R2 R2 Rs R5 1.) What is l1, the current that flows through the resistor R1? 2.) What is r, the internal resistance of the battery? 3.) What is l3, th 4.) What is P2, the power dissipated in resistor R2? 5.) What is V2, the magnitude of the voltage across the resistor R2? 6.) Resistor R2 is now shorted out as shown. How does the magnitude of the voltage across the battery change? e current through resistor Ra?Explanation / Answer
R3 + R4 + R5 = 28+90+90 = 208 ohm
Req = R2 = 1/ 1/81 + 1/208 = 58.29 ohm
I1 = 11.58 V/58.29 ohm = 0.198 A
(2)
V= I1( r+R)
12 V = 0.198 ( 58.45 + r)
12-11.578 = 0.1981 r
r = 2.125 ohms
(3)
I3 = 0.1981 * 81/81+208 = 0.0555 A
(4)
P2 = i^2 R2
= ( 0.1981-0.055)^2 81 = 1.64 W
(5)
V2 = ( 0.1981-0.0555) 81 = 11.5506 V
(6)
voltage across the battery decreases
sso option (A) is correct answer
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