Certainly! Let's analyze and solve this problem step-by-step.
### Step 1: Understand the Problem
1. We have a battery with a voltage of 10 V and an internal resistance of 1 Ω.
2. This battery is connected to an external resistor of 12 Ω.
3. We are to find the equivalent resistance, total resistance, and the current through the circuit.
### Step 2: Calculate the Equivalent Resistance
When resistors are connected in parallel, the formula to find the equivalent resistance [tex]\( R_{eq} \)[/tex] is:
[tex]\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \][/tex]
Where [tex]\( R_1 \)[/tex] is the internal resistance of the battery (1 Ω) and [tex]\( R_2 \)[/tex] is the external resistance (12 Ω). Simplifying this:
[tex]\[ \frac{1}{R_{eq}} = \frac{1}{1} + \frac{1}{12} = 1 + 0.08333 \approx 1.08333 \][/tex]
Taking the reciprocal to find [tex]\( R_{eq} \)[/tex]:
[tex]\[ R_{eq} = \frac{1}{1.08333} \approx 0.9231\ \Omega \][/tex]
### Step 3: Calculate the Total Resistance in the Circuit
Since the total resistance in this case is just the equivalent resistance we calculated, we get:
[tex]\[ R_{total} = 0.9231\ \Omega \][/tex]
### Step 4: Calculate the Current Using Ohm's Law
Ohm's Law states that:
[tex]\[ V = I \cdot R \][/tex]
Rearranging this formula to solve for current [tex]\( I \)[/tex]:
[tex]\[ I = \frac{V}{R} \][/tex]
Where [tex]\( V \)[/tex] is the voltage of the battery (10 V) and [tex]\( R \)[/tex] is the total resistance we just calculated (0.9231 Ω). So,
[tex]\[ I = \frac{10}{0.9231} \approx 10.8333\ \text{Amps} \][/tex]
### Final Values
From our calculations, we find:
1. Equivalent Resistance: 0.9231 Ω
2. Total Resistance: 0.9231 Ω
3. Current through the circuit: 10.8333 Amps
These values correspond well with the given numerical result. Therefore, we have:
Equivalent Resistance: [tex]\( 0.9231\ \Omega \)[/tex]
Total Resistance: [tex]\( 0.9231\ \Omega \)[/tex]
Current through the Circuit: [tex]\( 10.8333\ \text{Amps} \)[/tex]
