A 77.9−Ω resistor is connected in parallel with a 166.5−Ω resistor. This parallel group is connected in series with a 25.3−Ω resistor. The total combination is connected across a 12.6-V battery. Find (a) the current and (b) the power dissipated in the 166.5−Ω resistor.

We can solve this problem by breaking it down step by step, using the principles of resistors in parallel and series, Ohm's law, and the power formula.

### Step 1: Calculate the equivalent resistance of the parallel combination.
For two resistors \( R_1 \) and \( R_2 \) connected in parallel, the equivalent resistance \( R_{parallel} \) is given by:

\[
\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}
\]

Given:
- \( R_1 = 77.9 \ \Omega \)
- \( R_2 = 166.5 \ \Omega \)

\[
\frac{1}{R_{parallel}} = \frac{1}{77.9} + \frac{1}{166.5}
\]

Let's calculate \( R_{parallel} \):

\[
R_{parallel} = \left( \frac{1}{77.9} + \frac{1}{166.5} \right)^{-1}
\]

### Step 2: Find the total resistance of the circuit.
The total resistance of the circuit is the sum of the equivalent resistance of the parallel group and the resistance of the series resistor \( R_3 \).

\[
R_{total} = R_{parallel} + R_3
\]

Given:
- \( R_3 = 25.3 \ \Omega \)

### Step 3: Calculate the total current using Ohm's law.
Ohm's law states:

\[
I_{total} = \frac{V}{R_{total}}
\]

Where:
- \( V = 12.6 \ \text{V} \) is the voltage of the battery.

### Step 4: Find the current through the 166.5−Ω resistor.
The current through the 166.5−Ω resistor \( I_{166.5} \) in the parallel group is proportional to the total current flowing through the parallel combination. Since current divides in parallel circuits, use the following relation:

\[
I_{166.5} = \frac{V_{parallel}}{166.5}
\]

Where \( V_{parallel} \) is the voltage across the parallel resistors, which is the same for both \( R_1 \) and \( R_2 \). We can find \( V_{parallel} \) from the total current and resistances.

### Step 5: Calculate the power dissipated in the 166.5−Ω resistor.
The power dissipated in a resistor is given by:

\[
P = I^2 R
\]

So for the 166.5−Ω resistor:

\[
P_{166.5} = I_{166.5}^2 \times 166.5
\]

Now, let's calculate these values step by step.### Results:
1. The equivalent resistance of the parallel combination \( R_{parallel} \) is approximately \( 53.07 \ \Omega \).
2. The total resistance of the circuit \( R_{total} \) is approximately \( 78.37 \ \Omega \).
3. The total current flowing through the circuit \( I_{total} \) is approximately \( 0.161 \ \text{A} \).
4. The current through the 166.5−Ω resistor \( I_{166.5} \) is approximately \( 0.0512 \ \text{A} \).
5. The power dissipated in the 166.5−Ω resistor \( P_{166.5} \) is approximately \( 0.437 \ \text{W} \).

Would you like more details or have any questions?

Here are 5 related questions:
1. How would the current change if the 25.3-Ω resistor was removed?
2. How does the current split between resistors in parallel?
3. What would happen to the power dissipated if the 166.5-Ω resistor was doubled?
4. How do we calculate the total resistance in a parallel circuit with more than two resistors?
5. Can we use another method to calculate power dissipated in a resistor?

**Tip:** In parallel circuits, voltage across each component remains the same, but current divides inversely based on resistance values.

Learn how to find the current and power dissipated in a 166.5-Ω resistor connected in parallel with a 77.9-Ω resistor and in series with a 25.3-Ω resistor. This detailed step-by-step guide uses Ohm's Law, the power formula, and resistor combination rules. Suitable for Grades 11-12 physics students.

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