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Two lamps in parallel: total resistance, branch current, power, energy

SPH3U Grade 11 Physics · Electricity and Magnetism

A utility trailer carries two LED work lamps wired in parallel across a 12.0 V battery: a bright one of 30.0 Ω and a dimmer one of 60.0 Ω. The pair is left on for 3.00 h while a load is sorted in the dark. Find the total resistance the battery sees, the total current it supplies, the current in the 30.0 Ω lamp, the power that lamp dissipates, and the energy it uses over the three hours.

Step 1 of 5 · solve for Total resistance

Wired side by side, the lamps give the current two paths, so the pair resists less than either one alone — 20.0 Ω, below even the 30.0 Ω branch. Nothing about the current is knowable until this is settled.

Rearranged for R_t
Rt=R1R2R1+R2R_{t} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}
Your values, in your units
Rt=(30 Ω)⋅(60 Ω)(30 Ω)+(60 Ω)R_{t} = \frac{\left( 30\ \text{Ω} \right) \cdot \left( 60\ \text{Ω} \right)}{\left( 30\ \text{Ω} \right) + \left( 60\ \text{Ω} \right)}
Answer
Rt=20R_{t} = 20

Carried onward at full precision, not this rounded figure.

Open the Two Resistors in Parallel solver →

Step 2 of 5 · solve for Current

The battery holds 12.0 V across the whole network, so Ohm's law on the total resistance gives the total current drawn.

Rearranged for I
I=VRI = \tfrac{V}{R}
20 Ωcarried from step 1
Your values, in your units
I=(12 V)(20 Ω)I = \tfrac{\left( 12\ \text{V} \right)}{\left( 20\ \text{Ω} \right)}
Answer
I=600I = 600

Carried onward at full precision, not this rounded figure.

Open the Ohm's Law solver →

Step 3 of 5 · solve for Current in branch 1

That current splits, and not evenly. The easier path takes more, so the 30.0 Ω lamp gets two thirds of it — note the formula puts R₂ on top, which is the part everyone gets backwards.

Rearranged for I₁
I1=ItR2R1+R2I_{1} = I_{t} \frac{R_{2}}{R_{1} + R_{2}}
600 mAcarried from step 2
Your values, in your units
I1=(0.6 A)(60 Ω)(30 Ω)+(60 Ω)I_{1} = \left( 0.6\ \text{A} \right) \frac{\left( 60\ \text{Ω} \right)}{\left( 30\ \text{Ω} \right) + \left( 60\ \text{Ω} \right)}
Answer
I1=400I_{1} = 400

Carried onward at full precision, not this rounded figure.

Open the Current Divider solver →

Step 4 of 5 · solve for Power

Power in one element, from that element's own current and its own resistance. Feeding in the total current from step 2 here would be the classic blunder.

Rearranged for P
P=I2RP = I^{2} R
400 mAcarried from step 3
Your values, in your units
P=(0.4 A)2 (30 Ω)P = \left( 0.4\ \text{A} \right)^{2} \, \left( 30\ \text{Ω} \right)
Answer
P=4.8P = 4.8

Carried onward at full precision, not this rounded figure.

Open the Electrical Power (P = I²R) solver →

Step 5 of 5 · solve for Energy

Power held for three hours is energy. The answer lands in joules; a battery's amp-hour label is the same idea in disguise.

Rearranged for E
E=PtE = P t
4.8 Wcarried from step 4
Your values, in your units
E=(4.8 W) (3 h)E = \left( 4.8\ \text{W} \right) \, \left( 3\ \text{h} \right)
Converted to base units
E=(4.8 W) (10,800 s)E = \left( 4.8\ \text{W} \right) \, \left( 10{,}800\ \text{s} \right)
Answer
E=51.8E = 51.8

Carried onward at full precision, not this rounded figure.

Open the Electrical Energy (E = Pt) solver →

Why this order

Parallel is where the intuition built on series circuits breaks. Adding a second lamp lowers the total resistance and raises the current the battery must supply, which is the opposite of what most students predict and exactly why household outlets are wired this way: every device sees the full supply voltage and can be switched independently, at the cost of a service panel that has to be sized for the sum of everything switched on. Step 1 has to lead because the total resistance is the only thing that fixes the total current, and step 3 has to follow because power in one element depends on that element's own current, not the network's.

Step 3 is worth slowing down over. The current divider puts the other branch's resistance in the numerator, which looks wrong until you check it against the physics: the low-resistance path should carry the larger share, so the 30.0 Ω lamp takes 60/90 = two thirds of the 0.600 A, or 0.400 A, while the 60.0 Ω lamp takes 0.200 A. Those add back to 0.600 A, which is Kirchhoff's current law and a free check on the arithmetic. There is a second check available too: because both lamps sit across the same 12.0 V, P = V²/R gives 144/30 = 4.80 W directly, and it agrees with the I²R route in step 4. Over three hours that lamp costs 51.8 kJ — about 14 watt-hours, or roughly one percent of a small trailer battery.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.