Question

Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?​

Answer 1

Explanation:

The change in the current flowing through the electrical component can be determined by Ohm’s Law.

According to Ohm’s Law, the current is given by

↪I = V/R $\red\bigstar$

Now, the potential difference is reduced to half keeping the resistance constant,

↪Let the new voltage be V’ = V/2

↪Let the new resistance be R’ = R and the new amount of current be I’.

The change in the current can be determined using Ohm’s law as follows:

$\sf \leadsto \: I' = \frac{V'}{R'}$

$\leadsto \sf \: I' = \frac{ \frac{V'}{2} }{R'}$

$\leadsto \sf \: I' = \frac{1V'}{2R'}$

$\leadsto \sf \: I' = \frac{1}{2} \: \: \green \bigstar$

Therefore, the current flowing the electrical component is reduced by half.

Answer 2

Question :

Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?​

Given :

• Resistance will remain constant
• Potential difference b/w two ends will be half of the former value.

To find :

• What change will occur in the current through it

Solution :

Let the new current be I" and new potential difference be (V/2) and resistance will remain constant i.e. R

Using Ohm's law,

⇒ V = I * R

⇒ I = V/R  ................... (i)

Where,

• V refers to potential difference
• I refers to current
• R refers to resistance

Now substituting new values,

⇒ V/2 = I" * R

⇒ I" = (V/2)/R

⇒ I" = V/2R

⇒ I" = (1/2) V/R .................. (ii)

Now from equation (i) and (ii) we get,

∴ Current flowing in it becomes half of former value.