An ideal battery sends a current of 5 A in a resistor- When another resistor of value 10- is connected in parallel- the current through the battery is increased to 6 A- find the resistance of the first resistor-

R_1 = R, i_1 = 5A R_2 = 10R10+R, i_2 = 6A Since potential constant, i_1R_1 =i_2R_2 ⇒ 5 × R = 6× 10R10 + R ⇒ (10 + R)5 = 60 ⇒ 5R ...

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Standard X

Physics

Resistivity

An ideal batt...

Question

An ideal battery sends a current of $5 A$ in a resistor. When another resistor of value $10 Ω$ is connected in parallel, the current through the battery is increased to $6 A$ . find the resistance of the first resistor.

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Solution

$R_{1} = R, i_{1} = 5 A$

$R_{2} = \frac{10 R}{10 + R}, i_{2} = 6 A$

Since potential constant,
$i_{1} R_{1} = i_{2} R_{2}$

$\Rightarrow 5 \times R = \frac{6 \times 10 R}{10 + R}$

$\Rightarrow (10 + R) 5 = 60$

$\Rightarrow 5 R = 10 \Rightarrow R = 2 Ω$

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An ideal battery sends a current of 5A in a resistor. When another resistor of value 10Ω is connected in parallel, the current through the battery is increased to 6A. find the resistance of the first resistor.

R1=R, i1=5AR2=10R10+R,i2=6ASince potential constant,i1R1=i2R2⇒5×R=6×10R10+R⇒(10+R)5=60⇒5R=10⇒R=2Ω

An ideal battery sends a current of $5 A$ in a resistor. When another resistor of value $10 Ω$ is connected in parallel, the current through the battery is increased to $6 A$ . find the resistance of the first resistor.

$R_{1} = R, i_{1} = 5 A$

$R_{2} = \frac{10 R}{10 + R}, i_{2} = 6 A$

Since potential constant,
$i_{1} R_{1} = i_{2} R_{2}$

$\Rightarrow 5 \times R = \frac{6 \times 10 R}{10 + R}$

$\Rightarrow (10 + R) 5 = 60$

$\Rightarrow 5 R = 10 \Rightarrow R = 2 Ω$