Physics problems?

Question

A 76.0ohm & a 48.0ohm resistor are connected in parallel. When this combination is connected across a battery, the current delivered by the battery is 0.350A. When the 48.0ohm is resistor is disconnected, the current from the battery drops to 0.200A. Determine the emf of the battery.

2) Two cylindrical rods, one copper & the other iron, are identical in lengths and cross-sectional areas. They are joined, end-to-end, to form one long rod. A 30V battery is connected across the free ends of the copper-iron rod. What is the voltage between the ends of the copper rod?

Answer 1

1R = 1/76 + 1/48 = (48 + 76) / (48 X 76) = 124 / 3648 = 0.034

And R = 29.419 ohms

Current flowing through is 0.350 A

V = I.R = 0.35 X 29.419 = 10.297 V

When the 48 Ohms is removed, we are left with a resistance of 76 ohms and the current is 0.2 A.

V = 0.2 X 76 = 15.2 V

Why the difference in values? Due to the internal resistance of the battery. Let it be r ohms.

So, we have (29.419 + r) X 0.350 = (76 + r) X 0.2

10.297 + 0.35r = 15.2 + 0.2r

or 0.15 r = 15.2 - 10.3 = 4.9

r = 4.9 / 0.15 = 32.667 Ohms

The voltage is actually about 21.7 V but for the self resistance.

About the second question, since the conductivity of copper is very high compared to iron, most of the resistance in the circuit is contributed by iron and we can say that the voltage across the ends of the copper rod is almost zero (just as we neglect the resistance of the conducting wires or leads in the circuit).

Answer 2

You really need to quit screwing around and learn Ohms law ☺
1. The battery emf is 15.2 volts

2. You have to know the conductivity of each material to solve this problem.

Doug

Answer 3

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