#### Three resistors, R1 =10Ω, R2 =20Ω, and R3 =30Ω are connected in series to a 12 V battery.

#### Find:

a. The equivalent resistance of the circuit.

b. The current in each resistor.

c. The voltage across each resistor.

d. The power lost in each resistor.

To solve the problem step-by-step, we’ll look at the series circuit consisting of three resistors: R_{1} = 10 , \Omega, R_{2} = 20 , \Omega, R_{3} = 30 , \Omega, connected to a 12 V battery.

Equivalent Resistance

The equivalent resistance of resistors in series is found by simply adding their resistances together:

R_{t} =R_{1}+R_{2}+R_{3}

R_{t} =10Ω+20Ω+30Ω=60Ω

Therefore, the equivalent resistance of the circuit is 60 Ω.

Current in Each Resistor

Since the resistors are in series, the same current flows through each of them. We can calculate the total current (I) using Ohm’s law:

Thus, the current in each resistor is 0.2 A.

Voltage Across Each Resistor

The voltage across each resistor can be found using Ohm’s law (V = I \times R).

For R_{1}:

V_{1}=I×R_{1}=0.2A×10Ω=2V

For R_{2}:

V_{2}=I×R_{2}=0.2A×20Ω=4V

For R_{3}:

V_{3}=I×R_{3}=0.2A×30Ω=6V

Therefore, the voltages across the resistors are 2 V, 4 V, and 6 V, respectively.

Power Lost in Each Resistor

Finally, to calculate power (P) dissipated in each resistor, we use the formula:

P=I_{3}×R

For R₁:

P_{1}=(0.2A)2×10Ω=0.4W

For R_{2}:

P_{2}=(0.2A)2×20Ω=0.8W

For R_{3}:

P_{3}=(0.2A)2×30Ω=1.2W

The power lost in each resistor is 0.4 W, 0.8 W, and 1.2 W respectively.

**Examples & Evidence:**

An example of how this applies would be measuring the voltage across a light bulb and a heater in a series circuit and using the same formulas to find current and power in those components.

Ohm’s law and the formulas for equivalent resistance, voltage, and power are standard principles in electrical engineering and physics.

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