Question

The current \(I\) flowing through the given circuit will be ________A.

Answer 1

Correct Answer: 2

To find the current \( I \) flowing through the circuit, we use Ohm’s law: \( V = IR \). First, determine the total resistance in the circuit.
There are three resistors, each \( 9 \, \Omega \), in parallel. The formula for total resistance \( R_{\text{total}} \) for resistors in parallel is:
\( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \). 
Since \( R_1 = R_2 = R_3 = 9 \, \Omega \), we have:
\( \frac{1}{R_{\text{total}}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \).
Thus, \( R_{\text{total}} = 3 \, \Omega \).
Using Ohm's Law \( V = IR \), rearrange to find \( I \):
\( I = \frac{V}{R} = \frac{6}{3} = 2 \, \text{A} \).
The current \( I \) is \( 2 \, \text{A} \). This value falls within the expected range of 2 to 2, confirming our solution.

Answer 2

All 9 Ω resistances are in parallel.
R_eq\( = 3 Ω\)
\(I = \frac{6}{3} \, \text{A}\)
\(I=2A\)
So, the answer is 2A.