35 The diagram shows a cell of electromotive force (e.m.f.) 3.0 V and internal resistance 4.7 connected across a lamp. The lamp has a resistance of 9.3 .4.7 9.3 3.0 VWhat is the power dissipated by the internal resistance of the cell?A 0.22 W B 0.43 W C 0.64 W D 1.0 W36 A circuit consists of a battery, a high-resistance voltmeter and four fixed resistors, as shown. Thebattery has an electromotive force (e.m.f.) of 15.0 V and negligible internal resistance.V24.0 6.0 6.0 9.0 15.0 VWhat is the reading on the voltmeter?A 3.0 V B 6.0 V C 9.0 V D 12.0 V
Question
35 The diagram shows a cell of electromotive force (e.m.f.) 3.0 V and internal resistance 4.7 connected across a lamp. The lamp has a resistance of 9.3 .4.7 9.3 3.0 VWhat is the power dissipated by the internal resistance of the cell?A 0.22 W B 0.43 W C 0.64 W D 1.0 W36 A circuit consists of a battery, a high-resistance voltmeter and four fixed resistors, as shown. Thebattery has an electromotive force (e.m.f.) of 15.0 V and negligible internal resistance.V24.0 6.0 6.0 9.0 15.0 VWhat is the reading on the voltmeter?A 3.0 V B 6.0 V C 9.0 V D 12.0 V
Solution
Sure, let's solve these problems one by one.
Problem 35: First, we need to find the total current in the circuit. We can use Ohm's law for this, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). In this case, the total resistance is the sum of the internal resistance and the lamp resistance, which is 4.7 Ω + 9.3 Ω = 14 Ω. The voltage is the e.m.f., which is 3.0 V. So, the current is I = V/R = 3.0 V / 14 Ω = 0.214 A.
The power (P) dissipated by the internal resistance can be found using the formula P = I^2 * R, where I is the current and R is the resistance. Substituting the values we have, P = (0.214 A)^2 * 4.7 Ω = 0.214 A * 0.214 A * 4.7 Ω = 0.22 W. So, the answer is A, 0.22 W.
Problem 36: The circuit is a series circuit, so the total resistance is the sum of all the resistances, which is 24 Ω + 6 Ω + 6 Ω + 9 Ω = 45 Ω. The total voltage is the e.m.f., which is 15.0 V. So, the total current is I = V/R = 15.0 V / 45 Ω = 0.333 A.
The voltmeter is connected across the 6 Ω resistor. The voltage across this resistor can be found using Ohm's law, V = I * R. Substituting the values we have, V = 0.333 A * 6 Ω = 2.0 V. However, this is not one of the options. There might be a mistake in the problem or the options.
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