Consider the circuit shown below. Take the emf of the battery to be V_0 = 5V and

Question

Consider the circuit shown below. Take the emf of the battery to be V_0 = 5V and R = 10 Ohm. Assume that the wires are perfect conductors and that the battery has no internal resistance. Find the current in the circuit. Find the power generated by the battery. Find the power dissipated in the resistor. Consider again the circuit of problem 3. Again, take the emf of the battery to be V_0 = 5 V and R = 10 Ohm. But in this problem take the internal resistance of the battery to be 0.15 Ohm and the two wires that connect the battery to the resistor to each be 14 gauge gold wire of length 1 m. Find the current in the circuit. Find the power generated by the battery. Find the power dissipated in the resistor. Find the power dissipated in the wires. Find the power dissipated in the internal resistance of the battery. Find the terminal voltage across the battery.

Explanation / Answer

A) current i = Vo/R = 5 /(10+0.15+0.008+0.008) = 0.491 A

B) Power generated by battery is P = Vo*I = 5*0.491 = 2.455 W

C) Power dissipated in resistor is P = i^2*R = 0.491^2*10 = 2.41 W

D) power dissipated in wires is P = i^2*r = 0.491*2*0.008 = 0.007856 W

E) P = i^2*R = 0.491^2*0.15 = 0.03616 W

F) terminal voltage is V = e-(i*r) = 5-(0.491*0.15) = 4.93 V

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