Circuit question with two batteries and resistors in pa

In summary, the question asks about the power dissipated by resistor R1. The equations used to solve this problem are Ohm's law and the power equations. The current through R1 is determined by the PD across it, which can be found by using Kirchoff's circuit and voltage laws. The attempt at a solution involved simplifying the bottom resistor pair and setting up simultaneous equations to solve for the current. However, the resulting current of 1.05A seemed incorrect, leading to further questioning of the problem.
  • #1
FelaKuti
19
0

Homework Statement


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Q: What is the power dissipated by R1?

Homework Equations



Ohm's law: V = IR
Power: P = VI, I2R, V2/R
Kirchoff's circuit and voltage laws

The Attempt at a Solution



I assumed the currents were I1 traveling rightwards from the top battery, I2 rightwards from the second battery, and came together to make I3 for the bottom branch. The bottom resistor pair I simplified to 10/7 Ohms. From here I assume I have to set up simultaneous equations to solve for I1.

I used the equation for the bottom branch 1.5V - 10/7 * I3 = 0 to get I3 = 1.05A.

I subbed this into the equation for the outer loop 1.5 - 3I1 - 10/7I3 = 0. However plugging 1.05A into this gave me 0 for I1...

I feel like I'm way off the mark..
 
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  • #2
Why would there be any current flowing in R1?
 
  • #3
oz93666 said:
Why would there be any current flowing in R1?
Indeed!

The current through R1 is determined by the PD across R1. So ask yourself: what is the P.D. across R1?
 
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1. How do I calculate the total resistance in a circuit with two batteries and resistors in parallel?

To calculate the total resistance in a parallel circuit, you can use the formula 1/Rt = 1/R1 + 1/R2 + 1/R3 + ..., where Rt is the total resistance and R1, R2, R3, etc. are the individual resistances. In the case of two batteries and resistors in parallel, you would simply add the inverse of the two resistances together. For example, if R1 = 10 ohms and R2 = 20 ohms, the total resistance would be 1/Rt = 1/10 + 1/20 = 3/20, so Rt = 20/3 ohms.

2. How do I calculate the current in a specific branch of a parallel circuit?

To calculate the current in a specific branch of a parallel circuit, you can use Ohm's law, which states that I = V/R, where I is the current, V is the voltage, and R is the resistance. First, calculate the total resistance of the circuit using the formula mentioned in question 1. Then, use the voltage and the total resistance to calculate the current in the desired branch.

3. What is the purpose of using two batteries in a parallel circuit?

The purpose of using two batteries in a parallel circuit is to increase the total voltage in the circuit. When batteries are connected in parallel, the voltage remains the same, but the capacity (or amount of charge) increases. This can be useful in situations where a higher voltage is needed to power a device.

4. Can I add more resistors in parallel to a circuit with two batteries and resistors?

Yes, you can add more resistors in parallel to a circuit with two batteries and resistors. The total resistance would be calculated using the formula mentioned in question 1, where you would add the inverse of the new resistor to the existing total resistance. Keep in mind that adding more resistors will decrease the total resistance and therefore increase the total current in the circuit.

5. How do I calculate the voltage drop across each resistor in a parallel circuit?

To calculate the voltage drop across each resistor in a parallel circuit, you can use Ohm's law again. First, calculate the total current in the circuit using the formula I = V/R, where V is the total voltage and R is the total resistance. Then, use the current and the individual resistances to calculate the voltage drop across each resistor. The sum of all voltage drops should equal the total voltage in the circuit.

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