Question

If resistance in the circuit is increased 4 times by keeping the potential difference same the current in the circuit is what

Answer 1

Answer:

Since , Resistance is inversely proportional to the current , Current will become 1 / 4 th ..

Explanation:

Answer 2

$\bold{RESISTANCE }$

Resistance is the obstruction offered in the path of current.

$\bold{POTENTIAL \: DIFFERENCE }$

Potential difference is the amount of bringing a unit positive test positive charge from one point to another.

$\bold{OHM'S \: LAW}$

By Ohm's law we know that cuurrent flowing in a conductor is directly potential difference applied across it ends provided that physical conditions and the temperature of conductor remain constant.

So,

Let resistance be R , potential difference be V abd current I.

By ohm's law,

$\boxed{V = IR}$

$\huge \boxed{I = \frac{V}{R}}$---(1)

$\bold{ACCORDING \: TO \: QUESTION}$

Resistance is increased by 4 so resistance$R_2 = 4R$

Potential difference is $V_2= V$

Si, current will be,

$\huge \boxed{I_2 = \frac{V_2}{R_2}}$ ------(2)

On dividing eq(2) by eq(1),

$\dfrac{I}{I_2} = \frac{V}{R} ÷\dfrac{V_2}{R_2}$

$\dfrac{I}{I_2} = \dfrac{V}{R} \times \dfrac{4R}{V}$

$\dfrac{I}{I_2} = \dfrac{ \cancel{V}}{\cancel{R}} \times \dfrac{4 \cancel{R}}{ \cancel{V}}$

$\dfrac{I}{I_2} = 4$

$I = 4I_2$

$\dfrac{I}{4} = I_2$

So, the current will become one fourth.