Explanation:
We are given a circuit with a 12 V battery and three resistors connected in parallel:
- R₁ = 6 kΩ
- R₂ = 4 kΩ
- R₃ = 2 kΩ
Since they’re in parallel, the voltage across each resistor is the same:
12 V.
We need to find the current through each resistor using Ohm’s Law:
\[
I = \frac{V}{R}
\]
But note: resistances are in
kΩ, so we’ll keep units consistent. Let’s convert kΩ to Ω or work in mA (milliamps) to avoid large numbers.
Let’s use
mA and
kΩ:
Ohm’s Law becomes:
\[
I\ (\text{in mA}) = \frac{V\ (\text{in V})}{R\ (\text{in kΩ})}
\]
Because:
\[
1\ \text{mA} = \frac{1\ \text{V}}{1\ \text{kΩ}}
\]
So:
- \( I_{R1} = \frac{12\ \text{V}}{6\ \text{kΩ}} = 2\ \text{mA} \)
- \( I_{R2} = \frac{12\ \text{V}}{4\ \text{kΩ}} = 3\ \text{mA} \)
- \( I_{R3} = \frac{12\ \text{V}}{2\ \text{kΩ}} = 6\ \text{mA} \)
Total current from the battery (I_T) is the sum of all branch currents (Kirchhoff’s Current Law):
\[
I_T = I_{R1} + I_{R2} + I_{R3} = 2 + 3 + 6 = 11\ \text{mA}
\]
Let me double-check:
- 2 mA + 3 mA = 5 mA; 5 mA + 6 mA = 11 mA
✔
All voltages are 12 V, resistors are correctly read, calculations are simple division — no mistakes.
Final Answer:
11 mA
Parent Tip: Review the logic above to help your child master the concept of parallel circuit diagram.
