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1. A 15.0 kΩ resistor is hooked up to a 45.0 V battery in a circuit with a switch.
a.) Draw a circuit diagram for the circuit described. Label all parts and values.
b.) What is the current flowing through the resistor?
c.) What is the power dissipated by the resistor?                                                                    2. A 10.0 Ω resistor is hooked up in series with an 8.0 Ω resistor followed by a 27.0 Ω resistor. The circuit is powered by a 9.0 V battery.
a.) Draw a labeled circuit diagram for the circuit described.
b.) Calculate the equivalent resistance.
c.) Calculate the voltage drop across each resistor in the circuit.                                                 3. A 12.0 V battery is hooked up with three resistors (R1, R2, R3) in parallel with resistances of 2.0 Ω, 5.0 Ω, and 10.0 Ω, respectively.
a.) Draw a labeled circuit diagram for the circuit described.
b.) Calculate the equivalent resistance.
c.) Calculate the current passing through each resistor in the circuit.

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Explanation:

(1). Given that,

Resistance [tex]R = 15.0\times10^{3}\Omega[/tex]

Voltage [tex]V = 45.0\ V[/tex]

Using ohm's law

[tex]V = I\times R[/tex]

(a). Draw a circuit

(b).The current is defined as:

[tex]I = \dfrac{V}{R}[/tex].....(I)

Here, V = voltage

R = resistance

I = current

Put the value of V and R in equation (I)

[tex]I=\dfrac{45.0}{15.0\times10^{3}}[/tex]

[tex]I=0.003\ A[/tex]

The current is 0.003 A.

(c). The power dissipated by the resistor will be

[tex]P=\dfrac{V^2}{R}[/tex]

[tex]P = \dfrac{(45.0)^2}{15000}[/tex]

[tex]P = 0.135\ \ W[/tex]

The power dissipated by the resistor will be 0.135 Watt.

(2). Given that,

Resistance [tex]R_{1} = 10.0 \Omega[/tex]

Resistance [tex]R_{2} = 8.0 \Omega[/tex]

Resistance [tex]R_{3} = 27.0 \Omega[/tex]

Voltage [tex]V = 9.0\ \ V[/tex]

(a). Draw a circuit

(b). The equivalent circuit will be

[tex]R_{eq}=R_{1}+R_{2}+R_{3}[/tex]

[tex]R_{eq}=10+8+27[/tex]

[tex]R_{eq}=45\ \Omega[/tex]

The current is defined as:

[tex]I = \dfrac{V}{R}[/tex].....(I)

Here, V = voltage

R = resistance

I = current

Put the value of V and R in equation (I)

[tex]I=\dfrac{9.0}{45.0}[/tex]

[tex]I=0.2\ A[/tex]

The current is 0.2 A.

(c). The voltage drop across each resistor in the circuit

The voltage drop across 10.0 ohm resistor,

[tex]V = I\times R[/tex]

[tex]V = 0.2\times 10[/tex]

[tex]V = 2\ \ volt[/tex]

The voltage drop across 8.0 ohm resistor,

[tex]V = I\times R[/tex]

[tex]V = 0.2\times 8.0[/tex]

[tex]V = 1.6\ \ volt[/tex]

The voltage drop across 27.0 ohm resistor,

[tex]V = I\times R[/tex]

[tex]V = 0.2\times 27.0[/tex]

[tex]V = 5.4\ \ volt[/tex]

The voltage drop across each resistor in the circuit is 2 V, 1.6 V and 5.4 V.

(3). Given that,

Resistance [tex]R_{1} = 2.0 \Omega[/tex]

Resistance [tex]R_{2} = 5.0 \Omega[/tex]

Resistance [tex]R_{3} = 10.0 \Omega[/tex]

Voltage [tex]V = 12.0\ V[/tex]

(a). Draw a circuit

(b). The equivalent resistance will be

[tex]\dfrac{1}{R_{eq}}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}[/tex]

[tex]\dfrac{1}{R_{eq}}=\dfrac{1}{2.0}+\dfrac{1}{5.0}+\dfrac{1}{10.0}[/tex]

[tex]\dfrac{1}{R_{eq}}=\dfrac{4}{5}[/tex]

[tex]R_{eq}=1.25\ \Omega[/tex]

The equivalent resistance will be 1.25 ohm.

(c). The current passing through each resistor in the circuit.

The current passing through 2.0 ohm resistor

[tex]I = \dfrac{V}{R_{1}}[/tex]

[tex]I=\dfrac{12.0}{2.0}[/tex]

[tex]I = 6\ A[/tex]

The current passing through 5.0 ohm resistor

[tex]I = \dfrac{V}{R_{1}}[/tex]

[tex]I=\dfrac{12.0}{5.0}[/tex]

[tex]I = 2.4\ A[/tex]

The current passing through 10.0 ohm resistor

[tex]I = \dfrac{V}{R_{1}}[/tex]

[tex]I=\dfrac{12.0}{10.0}[/tex]

[tex]I = 1.2\ A[/tex]

The current passing through each resistor in the circuit is 6 A, 2.4 A and 1.2 A.

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