Question

12. Two resistors of 4 ohm and 6 ohm are connected in parallel. The combination is connected
across a 6 V battery. Answer the questions that follow:

a) Define 1 watt.
b) Draw the circuit diagram to depict the above situation.
c) Find the effective resistance of the circuit.
d) What would be the total power dissipated by the circuit.
e) Find the current passing through the 4 ohm resistor.​

Answer 1

1. -The watt (symbol: W) is a unit of power. In the International System of Units (SI) it is defined as a derived unit of 1 joule per second, and is used to quantify the rate of energy transfer.
2. -An electrical circuit is a path or line through which an electrical current flows. The path may be closed (joined at both ends), making it a loop. A closed circuit makes electrical current flow possible. It may also be an open circuit where the electron flow is cut short because the path is broken
3. -We can easily figure out from potential difference method that the resistors of 21 Ω and of 7 Ω are in parallel and also resistor of 3 Ω and 6 Ω below it as well as the resistors of 6 Ω and 12 Ω below it are in parallel
5. Hence reducing the circuit as explained above we can calculate the equivalent resistance as
6. 5
7. 14
10. -Power Rule: P = I × V
11. If a current I flows through through a given element in your circuit, losing voltage V in the process, then the power dissipated by that circuit element is the product of that current and voltage: P = I × V.
12. -Since the 4 ohm and the 4 ohm are connected in series, they have the same current I1, which must be equal to the current of the battery. Using Ohms law we get, I=RV=88=1A. Hence, the current flowing through the resistor R1 (4 ohms) is 1A

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Answer 2

Explanation:

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