Question

In the given electric circuit find the current shown by ammeter A.
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Hint:

For mixed circuits, reduce the parallel resistances first, then add the series resistances. Always use Ohm's law \( I = \frac{V}{R} \) for final current.

Answer 1

Given resistances in the circuit:
\[ R_1 = 5 \, \Omega, \, R_2 = 2 \, \Omega, \, R_3 = 3 \, \Omega, \, R_4 = 1.5 \, \Omega, \, \text{and the voltage } V = 8 \, \text{V}. \] To find the current shown by the ammeter, we need to calculate the total resistance in the circuit. The resistors are connected in a combination of series and parallel.
1. First, calculate the equivalent resistance of the resistors $R_2$ and $R_3$, which are in parallel:
\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}. \] Therefore,
\[ R_{\text{eq}} = \frac{6}{5} = 1.2 \, \Omega. \] 2. Now, the equivalent resistance $R_{\text{eq}} = 1.2 \, \Omega$ is in series with $R_1 = 5 \, \Omega$ and $R_4 = 1.5 \, \Omega$. The total resistance $R_T$ is:
\[ R_T = R_1 + R_{\text{eq}} + R_4 = 5 + 1.2 + 1.5 = 7.7 \, \Omega. \] 3. Using Ohm's law, the current $I$ is given by:
\[ I = \frac{V}{R_T} = \frac{8}{7.7} \approx 1.04 \, \text{A}. \] \[ \boxed{I \approx 1.04 \, \text{A}}. \]