A 9.5 V battery supplies a 1.5 mA current to a circuit for 5.0 h .
The correct answer and explanation is:
Let’s analyze the problem step-by-step.
Problem Statement:
- Voltage, V=9.5 VV = 9.5 \, \text{V}
- Current, I=1.5 mA=1.5×10−3 AI = 1.5 \, \text{mA} = 1.5 \times 10^{-3} \, \text{A}
- Time, t=5.0 hours=5.0×3600=18000 secondst = 5.0 \, \text{hours} = 5.0 \times 3600 = 18000 \, \text{seconds}
What is asked?
Usually, problems like this want you to find either:
- Total charge supplied by the battery,
- Total energy delivered by the battery,
- Or power dissipated in the circuit.
I’ll cover the charge and energy supplied, as these are common.
1. Total Charge Supplied
Electric current is the rate of flow of charge: I=Qt ⟹ Q=I×tI = \frac{Q}{t} \implies Q = I \times t
Plugging in values: Q=(1.5×10−3 A)×(18000 s)=27 CQ = (1.5 \times 10^{-3} \, A) \times (18000 \, s) = 27 \, C
So, the battery supplies 27 Coulombs of charge over 5 hours.
2. Total Energy Supplied
Electrical energy supplied by a battery is: E=V×QE = V \times Q
Where VV is voltage, and QQ is charge. E=9.5 V×27 C=256.5 JoulesE = 9.5 \, V \times 27 \, C = 256.5 \, \text{Joules}
So, the battery supplies about 256.5 Joules of energy over 5 hours.
Explanation (300 words):
A 9.5-volt battery driving a current of 1.5 milliamperes (mA) through a circuit for 5 hours results in both a measurable quantity of electric charge flowing through the circuit and a corresponding amount of electrical energy delivered to the components of the circuit.
Current (II) is the flow rate of electric charge and is measured in amperes (A). Since 1 mA = 0.001 A, the current here is 1.5×10−3 A1.5 \times 10^{-3} \, A. The total time duration for which the current flows is 5 hours, but to use SI units consistently, this time is converted into seconds (5 hours × 3600 seconds/hour = 18,000 seconds).
The total charge QQ that passes through the circuit is found by multiplying the current by the time: Q=I×t=1.5×10−3×18000=27 CoulombsQ = I \times t = 1.5 \times 10^{-3} \times 18000 = 27 \, \text{Coulombs}
This means 27 coulombs of charge have flowed from the battery through the circuit during the 5-hour period.
Next, the electrical energy EE delivered by the battery depends on both the charge moved and the voltage provided by the battery. Voltage represents the potential difference or “push” that moves the charges through the circuit. Multiplying the voltage by the charge gives energy in joules (J): E=V×Q=9.5×27=256.5 JE = V \times Q = 9.5 \times 27 = 256.5 \, \text{J}
This amount of energy is used to power the devices connected in the circuit, possibly converted to heat, light, or mechanical work depending on the load.
Understanding these calculations is fundamental in electronics and physics as it helps in designing circuits that match battery capacities and ensuring components receive the correct energy for their operation without overloading or wasting power.