Calculating Current in Parallel Resistor
The problem asks for the current flowing through the \(8\ \Omega\) resistor in a circuit containing resistors in parallel and series connected to a battery.
Step 1: Calculate the Equivalent Resistance of Parallel Resistors
Two resistors, \(R_1 = 8\ \Omega\) and \(R_2 = 24\ \Omega\), are connected in parallel. Their equivalent resistance (\(R_p\)) is calculated using the formula:
\(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\)
Substituting the values:
\(\frac{1}{R_p} = \frac{1}{8\ \Omega} + \frac{1}{24\ \Omega}\)
To add these fractions, find a common denominator (24):
\(\frac{1}{R_p} = \frac{3}{24\ \Omega} + \frac{1}{24\ \Omega} = \frac{4}{24\ \Omega}\)
Therefore, the equivalent parallel resistance is:
\(R_p = \frac{24\ \Omega}{4} = 6\ \Omega\)
Step 2: Calculate the Total Circuit Resistance
The parallel combination (\(R_p = 6\ \Omega\)) is connected in series with another resistor \(R_3 = 14\ \Omega\). The total equivalent resistance (\(R_{total}\)) of the circuit is the sum of the series resistances:
\(R_{total} = R_p + R_3\)
\(R_{total} = 6\ \Omega + 14\ \Omega = 20\ \Omega\)
Step 3: Calculate the Total Current from the Battery
The total voltage supplied by the battery is \(V = 12\ V\). Using Ohm's law (\(I = V/R\)), the total current (\(I_{total}\)) flowing from the battery is:
\(I_{total} = \frac{V}{R_{total}}\)
\(I_{total} = \frac{12\ V}{20\ \Omega} = 0.6\ A\)
This total current flows through the series resistor (\(R_3\)) and then splits between the parallel resistors (\(R_1\) and \(R_2\)).
Step 4: Calculate the Voltage Across the Parallel Combination
The voltage drop across the parallel combination (\(V_p\)) can be found using Ohm's law, using the total current flowing into the parallel section and the equivalent resistance of the parallel section:
\(V_p = I_{total} \times R_p\)
\(V_p = 0.6\ A \times 6\ \Omega = 3.6\ V\)
This is the voltage across both the \(8\ \Omega\) and \(24\ \Omega\) resistors.
Step 5: Calculate the Current in the \(8\ \Omega\) Resistor
Now, apply Ohm's law to the \(8\ \Omega\) resistor (\(R_1\)) using the voltage across the parallel combination (\(V_p\)):
\(I_{R1} = \frac{V_p}{R_1}\)
\(I_{R1} = \frac{3.6\ V}{8\ \Omega} = 0.45\ A\)
The current in the \(8\ \Omega\) resistor is \(0.45\ A\).
Similar Questions
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Important Questions from Combination of Resistors — Series and Parallel
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Three resistor, each equal to 3 Ω are connected so as to form a triangle. The equivalent resistance between any two vertices of the triangle is:
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