Question

7.5 x 10-3 C of charge flows in a circuit in 1.5 seconds. Find the current in the circuit. What would be the work done in doing so if the potential difference between two terminals of the battery is 1.5V ?

Answer 1

Solution :

⏭ Given:

• Charge flow = 7.5 × 10$^{-3}$ C
• Time interval = 1.5s
• P.d. = 1.5V

⏭ To Find:

• Current flow in circuit
• Work done to move charge

⏭ Concept:

✏ Current is defined as rate of charge flow in other words it is ratio of charge flow to the time taken.

⏭ Formula:

✏ Formula of current is given by...

$\bigstar \: \boxed{ \sf{ \pink{ \large{I = \dfrac{Q}{t} }}}}\: \bigstar$

✏ Formula of work done in terms of charge and potential difference is given by...

$\bigstar \: \boxed { \sf{ \large{ \green{W = Q \times \triangle{V}}}}} \: \bigstar$

⏭ Calculation:

_________________________________

• Current flow

$\circ \sf \: I = \dfrac{7.5 \times {10}^{ - 3} }{1.5} \\ \\ \circ \: \boxed{ \purple{ \tt{I = 5 \: mA}}}$

_________________________________

• Work done

$\circ \sf \: W = 7.5 \times {10}^{ - 3} \times 1.5 \\ \\ \circ \: \boxed{ \sf{ \orange{W = 11.25 \: mJ}}}$

________________________________

Answer 2

$\mathtt{\huge{ \fbox{Solution :)}}}$

Given ,

Charge (q) = 7.5 x (10)^-3 C

Time (t) = 1.5 Sec

Potential difference (v) = 1.5 V

We know that , the rate of unit charge is called electric current

$\large \mathtt{ \fbox{Electric \: current \: (I) = \frac{q}{t} }}$

Thus ,

I = [7.5 × (10)^-3]/1.5

I = 5 × (10)^-3 amp

Hence , the current is 5 × (10)^-3 amp

Now , we have to find the work done when 7.5 x (10)^-3 C charge move from one point to another point

$\large \mathtt{\fbox{Potential \: difference \: (v) = \frac{work \: done}{charge} }}$

Thus ,

1.5 = work done/7.5 x (10)^-3

Work done = 11.25 × (10)^-3 Joule

Hence , the work done is 11.25 × (10)^-3 Joule

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