Question

Calculate the equivalent resistance of three resistors 8 Ω, 4 Q and 6 Ω connected in series. Also calculate the current if the combination is connected to a battery of 9 volt.



class 10
cbse
physics​

Answer 1

Here,

$\: \: \: \: { \dot \: \: \: R_1 = 8 Ω}$

$\: \: \: \: { \dot \: \: \: R_2= 4 Ω}$

$\: \: \: \: { \dot \: \: \: R_3 = 6 \: Ω}$

★ So, equivalent resistance (R)

$\bold{→ R_1 + R_2 + R_3}$

$\bold{→ 8Ω + 4 Ω+ 6 Ω}$

$\underline \bold{→1 8 \: Ω}$

Again this combination is connected to a battery of 9 volt.

So,

• V (Pot. difference) → 9 V
• I ( Current ) → ?

We know that ,

$\large { \underline{\color{red} {\bold{V = IR}}}}$

$\bold{→ 9 = I × 18}$

$\color{blue} \bold{→I = \frac{9}{18} }$

$\color{blue} \bold{→I = \frac{1}{2} }$

$\color{blue} \bold{→ \boxed {\large{\bold{I = 0.5A}}}}$