Question

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.

Answer 1

Please see attachment

Answer 2

Step-by-step explanation:

(1). Given that,

Resistance
`R = 15.0*10^(3)\Omega`

Voltage
`V = 45.0\ V`

Using ohm's law

`V = I* R`

(a). Draw a circuit

(b).The current is defined as:

`I = (V)/(R)`.....(I)

Here, V = voltage

R = resistance

I = current

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

`I=(45.0)/(15.0*10^(3))`

`I=0.003\ A`

The current is 0.003 A.

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

`P=(V^2)/(R)`

`P = ((45.0)^2)/(15000)`

`P = 0.135\ \ W`

The power dissipated by the resistor will be 0.135 Watt.

(2). Given that,

Resistance
`R_(1) = 10.0 \Omega`

Resistance
`R_(2) = 8.0 \Omega`

Resistance
`R_(3) = 27.0 \Omega`

Voltage
`V = 9.0\ \ V`

(a). Draw a circuit

(b). The equivalent circuit will be

`R_(eq)=R_(1)+R_(2)+R_(3)`

`R_(eq)=10+8+27`

`R_(eq)=45\ \Omega`

The current is defined as:

`I = (V)/(R)`.....(I)

Here, V = voltage

R = resistance

I = current

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

`I=(9.0)/(45.0)`

`I=0.2\ A`

The current is 0.2 A.

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

The voltage drop across 10.0 ohm resistor,

`V = I* R`

`V = 0.2* 10`

`V = 2\ \ volt`

The voltage drop across 8.0 ohm resistor,

`V = I* R`

`V = 0.2* 8.0`

`V = 1.6\ \ volt`

The voltage drop across 27.0 ohm resistor,

`V = I* R`

`V = 0.2* 27.0`

`V = 5.4\ \ volt`

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

(3). Given that,

Resistance
`R_(1) = 2.0 \Omega`

Resistance
`R_(2) = 5.0 \Omega`

Resistance
`R_(3) = 10.0 \Omega`

Voltage
`V = 12.0\ V`

(a). Draw a circuit

(b). The equivalent resistance will be

`(1)/(R_(eq))=(1)/(R_(1))+(1)/(R_(2))+(1)/(R_(3))`

`(1)/(R_(eq))=(1)/(2.0)+(1)/(5.0)+(1)/(10.0)`

`(1)/(R_(eq))=(4)/(5)`

`R_(eq)=1.25\ \Omega`

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

`I = (V)/(R_(1))`

`I=(12.0)/(2.0)`

`I = 6\ A`

The current passing through 5.0 ohm resistor

`I = (V)/(R_(1))`

`I=(12.0)/(5.0)`

`I = 2.4\ A`

The current passing through 10.0 ohm resistor

`I = (V)/(R_(1))`

`I=(12.0)/(10.0)`

`I = 1.2\ A`

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