Question

a charge of 5000 current flows through an electric circuit in 2 hours and 30 minutes calculate the magnitude of current in circuit​

Answer 1

Given, Charge(Q) = 5000 Coulomb

Time(t) = 2 hours and 30 minutes

Convert time into seconds:

So, 1 minute = 60 seconds

1 hour = 60 minute

Time(t) = 2 × 60 × 60 + 30 × 60

= 2 × 3600 + 1800

= 7200 + 1800

= 10800 seconds

we know, Current(i) = charge / time

i = 5000 / 10800

i = 0.46 ampere

Therefore, current i circuit is 0.46 A

Answer 2

Answer:

The magnitude of the current in the circuit is 0 . 56 A .

Step-by-step explanation:

Step 1: Given data:

1. The charge flowing through the circuit is $Q=5000 \mathrm{C}$

2. The time taken is,

$\begin{aligned}& \mathrm{t}=2 \text { hours } 30 \text { minutes } \\& \mathrm{t}=2 \times 60 \times 60+30 \times 60 \\& \mathrm{t}=9000 \text { seconds }\end{aligned}$

Step 2: To find the magnitude of current in the circuit:

We know that,

Current=Charge*

$\Rightarrow 5000/9000$

$\begin{aligned}& =59 \\& =0.56 \mathrm{~A}\end{aligned}$

Therefore, the magnitude of the current in the circuit is$0.56 \mathrm{~A}$.

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