Question

name: ireton stonemel period: honors simple series circuit worksheet 4 1. 120 volts 1. what is the total resistance of the circuit? 2. solve for current (i) 3. calculate the voltage drop across the 4 ohm resistor 4. what is the remaining voltage and current at c? v = i = 2. i = 4 amps 48 volts 5. what is the total resistance? 6. what is the resistance of x? 7. what is the voltage drop across the 6ω resistor? 8. what is the remaining voltage at d?

Explanation:

Step1: Calculate total resistance for first circuit

In a series - circuit, total resistance $R_T$ is the sum of individual resistances. For the first circuit with resistances $R_1 = 16\Omega$, $R_2=20\Omega$, $R_3 = 4\Omega$, $R_T=16 + 20+4=\ 40\Omega$.

Step2: Calculate current for first circuit

Using Ohm's law $I=\frac{V}{R}$, where $V = 120$ volts and $R = R_T=40\Omega$. So $I=\frac{120}{40}=3$ amps.

Step3: Calculate voltage drop across 4 - ohm resistor

Using Ohm's law $V = IR$, with $I = 3$ amps and $R = 4\Omega$. So $V=3\times4 = 12$ volts.

Step4: Find voltage and current at C for first circuit

In a series - circuit, current is the same throughout, so $I = 3$ amps at C. The voltage across the $16\Omega$ and $20\Omega$ resistors combined is $V=(16 + 20)\times3=108$ volts. So the voltage at C relative to the negative terminal of the battery is $120-108 = 12$ volts.

Step5: Calculate total resistance for second circuit

Using Ohm's law $R=\frac{V}{I}$, with $V = 48$ volts and $I = 4$ amps, $R_T=\frac{48}{4}=12\Omega$.

Step6: Calculate resistance of X for second circuit

Given resistances $R_1 = 3\Omega$ and $R_2=6\Omega$, and $R_T=12\Omega$. In a series - circuit $R_T=R_1+R_2+R_X$. So $R_X=12-(3 + 6)=3\Omega$.

Step7: Calculate voltage drop across 6 - ohm resistor for second circuit

Using Ohm's law $V = IR$, with $I = 4$ amps and $R = 6\Omega$. So $V=4\times6=24$ volts.

Step8: Calculate remaining voltage at D for second circuit

The voltage drop across the $3\Omega$ and $6\Omega$ resistors combined is $(3 + 6)\times4 = 36$ volts. So the remaining voltage at D relative to the negative terminal of the battery is $48-36 = 12$ volts.

Answer:

1. $40\Omega$
2. $3$ amps
3. $12$ volts
4. $V = 12$ volts, $I = 3$ amps
5. $12\Omega$
6. $3\Omega$
7. $24$ volts
8. $12$ volts