Find voltage and power given current and resistance.

Question

Circuit with two batteries, each with internal resistance, that are connected in series with each other and a resistor R. Battery 1 has emf of 6 V and r_1 of 1.3 ohms. Battery 2 (connected in the opposite direction as battery 1) has emf of 9 V and r_2 of 1.4 ohms.

a) Find the voltage drop across resistor R.

b) Find the power dissipated in R.

My answers are incorrect, but here's what I did. For a, I used V = I(r₁+r₂) and got 1.431 V. For b, I used P = IV and got 0.758 W. Did I make a mistake in the formulas or the calculations?

Answer 1

Given Data:

1. Battery 1 (emf E1)= 6 V, internal resistance r1_= 1.3 Ω
2. Battery 2 (emf E2​) = 9 V, internal resistance r2 = 1.4 Ω (connected in the opposite direction to Battery 1)
3. Resistor R is connected in series with both batteries.

Since Battery 2 is connected in the opposite direction to Battery 1, their emfs will partially oppose each other.

Step 1: Find the Net EMF

The net emf (Enet) in the circuit will be:

Enet=E1-E2=6V- 9V= -3V

Step 2: Find the Total Internal Resistance

The total internal resistance (Rint​) of the circuit is the sum of the internal resistances of both batteries:

Rint = r1+r2 = 1.3Ω+1.4Ω = 2.7Ω

Step 3: Apply Ohm’s Law to Find the Current

Let I be the current flowing through the circuit. Using Ohm’s Law:

I = Enet/(Rint+R)

I = -3V/(2.7Ω + R)

Problem Recap:

We need to find R in a circuit with a known net emf and internal resistance. However, we can't directly find the current because it depends on R, and finding R also depends on knowing the current. This creates a circular dependency.

Approach to Solve for R:

To solve for R without knowing the current directly, we express all relevant variables in terms of R and solve the resulting equation.

Step-by-Step Solution:

1. Set Up the Equation for Current: We know:

I=Enet/(R+Rint)

Substitute the given net emf (Enet= −3 V and total internal resistance Rint=2.7 Ω

I=-3/(2.7Ω+R)

1. Express the Voltage Drop Across R: The voltage drop across R is given by:

VR=I*R

Substitute I from above:

VR=(-3/(2.7Ω+R) *R

1. Express Power Dissipation: The power dissipated in R is:

P=I²*R

Substitute I:

P= [(-3/(R+2.7Ω)]² * R = 9R/(2.7Ω+R)²

Answer 2

Let's analyze the circuit with the two batteries in series and the resistor \( R \).

### a) Finding the voltage drop across resistor \( R \)

1. **Total EMF Calculation**:

- Battery 1: \( \text{EMF}_1 = 6 \, \text{V} \)

- Battery 2 (in opposite direction): \( \text{EMF}_2 = -9 \, \text{V} \)

The total EMF in the circuit:

\[

\text{Total EMF} = 6 \, \text{V} - 9 \, \text{V} = -3 \, \text{V}

\]

2. **Total Internal Resistance**:

\[

r_{\text{total}} = r_1 + r_2 = 1.3 \, \Omega + 1.4 \, \Omega = 2.7 \, \Omega

\]

3. **Calculating the Current**:

Using Ohm's law, the total voltage \( V \) across the internal resistances and the external resistor \( R \):

\[

V = I(r_{\text{total}}) + V_R

\]

Since we know the total EMF is -3 V, we can set up the equation:

\[

-3 = I(2.7) + V_R

\]

Rearranging gives:

\[

V_R = -3 - I(2.7)

\]

4. **Finding the Voltage Drop Across \( R \)**:

Since we don’t have \( I \) yet, let’s express \( I \) first. The current \( I \) can be calculated by rearranging the equation:

\[

I = \frac{-3 - V_R}{2.7}

\]

To find \( V_R \), we will set up the equation:

\[

-3 = I(2.7) + V_R

\]

Substituting \( V_R \) into the equation gives us a relationship that we can solve once we express \( V_R \) in terms of \( I \).

However, we can also express it as:

\[

I = \frac{-3}{2.7 + R}

\]

Now, substitute \( I \) back into the expression for \( V_R \):

\[

V_R = -3 - \left(\frac{-3}{2.7 + R}\right)(2.7)

\]

### b) Finding the power dissipated in \( R \)

The power dissipated in the resistor \( R \) can be calculated as:

\[

P = I^2 R

\]

Using the current found in part (a), substitute \( I \):

\[

P = \left(\frac{-3}{2.7 + R}\right)^2 R

\]

### Summary of Formulas:

- Current through the circuit:

\[

I = \frac{-3}{2.7 + R}

\]

- Voltage drop across \( R \):

\[

V_R = -3 - I(2.7)

\]

- Power in \( R \):

\[

P = I^2 R

\]

### Conclusion:

Your initial calculations seem to have gone wrong mainly in handling the total EMF and using it correctly. Be sure to substitute the values properly and compute using the derived expressions. If you have a specific value for \( R \), substitute it into the equations to find the exact voltage drop and power.