Question

Three resistors are joined in parallel. A current of 7.5 A flows due to 30 V battery. If two resistors have resistances of 10 Ω and 12 Ω. Find out the third. Also find the current passing through each resistor.

Answer 1

Let - the resistance of three resistors are R1 , R2 and R3 which are connected in parallel.

$\large \underline \bold{Given :-}$

$\: \: \: \: \: \:$$\sf{R1 = 10}$Ω

$\: \: \: \: \: \:$$\sf{R2 = 12}$Ω

$\sf{Current \: flow \: (I) = 7.5 \: A}$

$\: \: \: \:$$\sf{Voltage \: (V) = 30 \: v}$

$\large \underline \bold{To \: Find :-}$

$\sf{Calculate \: the \: resistance \: of \: third \: resistor \: ?}$

$\sf{also \: the \: current \: passing \: through \: all \: resistors \: ?}$

$\large \underline \bold{Using \: Formulas :-}$

When R1 , R2 and R3 are connected in parallel .

then , the equivalent resistance -

$\small \bold{\dfrac{1}{R} =\dfrac{1}{R1} + \dfrac{1}{R2} + \dfrac{1}{R3}}$

$\: \: \: \: \: \: \: \:$$\small \bold{Curret \: (I) =\dfrac{V}{R}}$

$\large \underline \bold{Solution :-}$

$\sf{On \: Using \: the \: above \: formula -}$

$\small \bold{\dfrac{1}{R} =\dfrac{1}{R1} + \dfrac{1}{R2} + \dfrac{1}{R3}}$

$\small \bold{R =\dfrac{R1\times R2\times R3}{R1 R2 + R2 R3 + R3 R1}}$

$\sf{R =\dfrac{10\times 12\times R3}{10(12) + 12R3 + 10R3}}$

$\sf{R =\dfrac{120R3}{(120 + 22R3)} ----(1)}$

Now ,

$\small \bold{resistance \: (R) =\dfrac{V}{I}}$

$\: \: \: \: \: \:$$\sf{R =\dfrac{30}{7.5}}$

$\: \: \: \: \: \:$$\sf{R = 4}$Ω$---(2)$

From eq.(1) & eq.(2) -

$\: \: \:$$\sf{4 =\dfrac{120R3}{(120 + 22R3)}}$

$\: \: \:$$\sf{1 =\dfrac{30R3}{(120 + 22R3)}}$

$\: \: \:$$\sf{120+ 22R3 = 30R3}$

$\: \: \:$$\sf{30R3 - 22R3 = 120}$

$\: \: \: \: \: \:$$\sf{8R3 = 120}$

$\: \: \: \: \: \: \: \:$$\sf{R3 =\dfrac{120}{8} = 15}$Ω

$\small \bold{resistance \: of \: third \: resistor \: is \: 15}$Ω.

$\small \bold{1) \: Current \: flow \: through \: first \: resistor -}$

$\: \: \:$$\small \bold{I1 = \dfrac{V}{R1} =\dfrac{30}{10} = 3 \: A}$

$\small \bold{2) \: Current \: flow \: through \: Second \: resistor -}$

$\: \: \:$$\small \bold{I2 = \dfrac{V}{R2} =\dfrac{30}{12} = 2.5 \: A}$

$\small \bold{3) \: Current \: flow \: through \: third \: resistor -}$

$\: \: \:$$\small \bold{I3 = \dfrac{V}{R3} =\dfrac{30}{15} = 2 \: A}$

Answer 2

Answer:

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