Question

a series circuit consisting of three resistors having 40 ohms ,50 ohms and 20 ohms respectively is connected across a voltage source of 120V.find the current and potential difference across each resistor.

Answer 1

Answer on Question 58755, Physics, Electric Circuits

Question:

A series circuit consisting of three resistors having $40\Omega$ , $50\Omega$ and $20\Omega$ respectively is connected across a voltage source of $120V$ . Find the current and potential difference across each resistor.

Solution:

Let's consider a series circuit consisting of three resistors having the resistances of $R_{1} = 40\Omega$ , $R_{2} = 50\Omega$ and $R_{3} = 20\Omega$ respectively. These three resistors are connected across a voltage source of $120V$ .

a) Let's first find the equivalent resistance of resistors in series:

$$R _ {e q} = R _ {1} + R _ {2} + R _ {3} = 4 0 \Omega + 5 0 \Omega + 2 0 \Omega = 1 1 0 \Omega .$$

As we know, in series circuit the current is the same through each resistor. Then, from the Ohm's law we can find the current in this circuit:

$$I = \frac {V}{R _ {e q}} = \frac {V}{R _ {1} + R _ {2} + R _ {3}} = \frac {1 2 0 V}{4 0 \Omega + 5 0 \Omega + 2 0 \Omega} = \frac {1 2 0 V}{1 1 0 \Omega} = 1. 0 9 A.$$

b) Then, we can find the potential difference (or voltage drops) across each resistors:

$$V _ {1} = I R _ {1} = 1. 0 9 A \cdot 4 0 \Omega = 4 3. 6 \mathrm {V},$$ $$V _ {2} = I R _ {2} = 1. 0 9 A \cdot 5 0 \Omega = 5 4. 5 \mathrm {V},$$ $$V _ {3} = I R _ {3} = 1. 0 9 A \cdot 2 0 \Omega = 2 1. 8 \mathrm {V}.$$

Let's check our calculations. As we know, in the series circuit the total potential difference across the resistors is equal to the potential difference across the voltage source:

$$V = V _ {1} + V _ {2} + V _ {3},$$ $$120 \, V = 43.6 \, V + 54.5 \, V + 21.8 \, V,$$ $$120 \, V = 120 \, V.$$

Therefore, we do all the calculations correctly.

Answer:

a) $I = 1.09 \, A$.

b) $V_1 = 43.6 \, V, V_2 = 54.5 \, V, V_3 = 21.8 \, V$.

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