For the circuit diagram of the value current through 6 - resistor and p-d across the 12 - resistor

R_1=9Ω R_2=15Ω 4= I_1 R_1=I_1× 9 I_1=49A 4= I_2 R_2=I_2× 15 I_2=415A Potential difference for 12 Ω = I_2× 12=415× 12=3.2V ...

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Byju's Answer

Standard XII

Physics

Kirchhoff's Junction Law

for the circu...

Question

for the circuit diagram of the value current through $6 Ω$ resistor and $p . d$ across the $12 Ω$ resistor

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Solution

$R_{1} = 9 Ω$
$R_{2} = 15 Ω$
4= $I_{1} R_{1} = I_{1} \times 9$
$I_{1} = \frac{4}{9} A$
4= $I_{2} R_{2} = I_{2} \times 15$
$I_{2} = \frac{4}{15} A$
Potential difference for 12 $Ω$ = $I_{2} \times 12 = \frac{4}{15} \times 12 = 3.2 V$

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for the circuit diagram of the value current through 6 Ω resistor and p.d across the 12 Ω resistor

R1=9ΩR2=15Ω4=I1R1=I1×9I1=49A4=I2R2=I2×15I2=415APotential difference for 12 Ω=I2×12=415×12=3.2V

for the circuit diagram of the value current through $6 Ω$ resistor and $p . d$ across the $12 Ω$ resistor

$R_{1} = 9 Ω$
$R_{2} = 15 Ω$
4= $I_{1} R_{1} = I_{1} \times 9$
$I_{1} = \frac{4}{9} A$
4= $I_{2} R_{2} = I_{2} \times 15$
$I_{2} = \frac{4}{15} A$
Potential difference for 12 $Ω$ = $I_{2} \times 12 = \frac{4}{15} \times 12 = 3.2 V$