The given circuit consists of a battery of 5V and several resistors arranged in a mixed configuration. We will find the total current supplied by the battery by analyzing the circuit's resistance.
Identify Resistors in Series and Parallel: The three 5 Ω resistors and one 2.5 Ω resistor are involved.
The two 5 Ω resistors at the top are in parallel.
The equivalent resistance (\(R_p\)) of two 5 Ω resistors in parallel is given by:
\(R_p = \frac{1}{\frac{1}{5}+\frac{1}{5}} = \frac{1}{0.4} = 2.5 Ω\)
Calculate Total Resistance: The equivalent parallel resistance (\(R_p = 2.5 Ω\)) is in series with the other 5 Ω resistor and the 2.5 Ω resistor.
Use Ohm’s Law to Find Current: Ohm’s Law states \(V = IR\). Thus, the current (\(I\)) is:
\(I = \frac{V}{R} = \frac{5 V}{10 Ω} = 0.5 A\)
Verify Solution Within Range: The range provided is 2,2. Our calculated current \(0.5 A\) is not within this range, indicating a possible misunderstanding or error in the expected range setup.
The total current supplied to the circuit by the 5V battery is 0.5 A.
