An electric device draws 5.60 A at 240 V. If the resistance of the device were reduced by 35 % , what current would be drawn at 240 V ?
The correct answer and explanation is:
✅ Correct Answer:
8.62 A
🔍 Explanation (300 words):
To solve this problem, we apply Ohm’s Law, which states: V=IRorI=VRV = IR \quad \text{or} \quad I = \frac{V}{R}
Where:
- VV is the voltage (in volts),
- II is the current (in amperes),
- RR is the resistance (in ohms).
Step 1: Find the original resistance
We are given:
- Original current, I1=5.60 AI_1 = 5.60 \, \text{A}
- Voltage, V=240 VV = 240 \, \text{V}
Using Ohm’s Law: R1=VI1=2405.60=42.857 ΩR_1 = \frac{V}{I_1} = \frac{240}{5.60} = 42.857 \, \Omega
Step 2: Calculate the new resistance
The resistance is reduced by 35%, so the new resistance is: R2=R1×(1−0.35)=42.857×0.65=27.857 ΩR_2 = R_1 \times (1 – 0.35) = 42.857 \times 0.65 = 27.857 \, \Omega
Step 3: Calculate the new current
Now using Ohm’s Law again: I2=VR2=24027.857≈8.62 AI_2 = \frac{V}{R_2} = \frac{240}{27.857} \approx 8.62 \, \text{A}
✅ Final Answer: 8.62 amperes
⚡ Summary:
The original current was 5.60 A at 240 V. When the resistance is reduced by 35%, the new resistance allows more current to flow because current and resistance are inversely related (as resistance goes down, current goes up). By recalculating using Ohm’s Law, we find that the new current drawn is 8.62 A, a significant increase due to the lower resistance in the circuit.