Electricity - Series and Parallel Circuit

AI Thread Summary
The discussion centers on a circuit with a 22-ohm and a 33-ohm resistor connected in series across a 120-V source. The equivalent resistance is calculated to be 55 ohms, resulting in a circuit current of 2.2 A. The voltage drop across the 22-ohm resistor is 48.4 V, while the drop across the 33-ohm resistor is 70.4 V. The total voltage drop across both resistors equals the source voltage of 120 V, confirming Kirchhoff's second law. The calculations demonstrate the relationships between resistance, current, and voltage in series circuits.
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Homework Statement



A 22-ohm resistor and a 33-ohm resistor are connected in series and placed across 120-V potential difference.

1. what is the equivalent resistance of the circuit?

2. what is the current in the circuit?

3. what is the voltage drop across each of resistor?

4. What is the voltage drop across the two resistors together?

Homework Equations



R = RA + RB
I = V source / RA + RB
VA = IRA
VB = IRB

The Attempt at a Solution



1. what is the equivalent resistance of the circuit?
R = RA + RB
R = 22-ohm + 33-ohm
R = 55-ohm

2. what is the current in the circuit?
I = Vsource/ RA + RB
I = 120/ 55 = 2.2 A

3. what is the voltage drop across each of resistor?

VA = IRA
= 48.4 V

VB = IRB
= 70.4 V

4. What is the voltage drop across the two resistors together?
- Ok, they are asking for the voltage drop across the two resistors together. I think you would have to add the two resistors but then should i divide it by the number of volts. But I know you would have to add the total of resistors.
 
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Yes you would just add them. Ideally, the sum voltage drops would be the same as the source voltage. (Kirchoff's 2nd law for circuits)
 
Recalculate I*RB.
The voltage drop across the two resistors together is the sum of voltages on each resistor.

ehild
 
Yes to ^. With the source voltage being the 120 V.
 
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