A circuit is constructed with five resistors and one real battery as shown above

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 = 580, R4-R5-130 and R2-830. The measured voltage across the terminals of the batery is battery = 11.48 V I, R I2 R I2 R 2 1) What is l1, the current that flows through the resistor R1? mA Submit 2) What is r, the internal resistance of the battery? Submit 3) What is 13, the current through resistor R3? mA Submit 4) What is P2, the power dissipated in resistor R2? Submit 5) What is V2, the magnitude of the voltage across the resistor R2? Submit

Explanation / Answer

1) R345 = R3 + R4 + R5

= 58 + 130 + 130

= 318 ohms

R2345 = R2*R345/(R2 + R345)

= 83*318/(83 + 318)

= 65.8 ohms

R12345 = R1 + R2345

= 58 + 65.8

= 123.8 ohms

I1 = V_terminal/R12345

= 11.48/123.8

= 0.09273 A

= 92.7 mA

2) we know, V_terminal = V_emf - I1*r

==> r = (V_emf - V_terminal)/I

= (12 - 11.48)/0.09273

= 5.61 ohms

3) I3 = I1*R2/(R2 + R345)

= 92.7*83/(83 + 318)

= 19.2 mA

4) I2 = I1 - I3

= 92.7 - 19.2

= 73.5 mA

Power dissipated at R2, P2 = I2^2*R2

= (73.5*10^-3)^2*83

= 0.448 W

5) V2 = I2*R2

= 73.5*10^-3*83

= 6.10 V

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