Question

a bulb is connected in series to 20 volt battery the circuit resistance is 5 ohm and the current of 2 ampere flows through the circuit calculate the resistance of bulb now a resistance of 10 ohm added in parallel to the existing circuit then what is the change in current flowing through the original 5 ohm circuit​

Answer 1

resistance of the bulb= $\frac{V}{I}$

                                 = 20/2

                                   = 10 Ω

NEW CURRENT FLOWING THROUGH THE CIRCUIT AFTER ADDING THE 10Ω RESISTOR

TOTAL RESISTANCE OF THE CIRCUIT

$\frac{1}{R}$   =  $\frac{1}{10}$    + .   $\frac{1}{10}$

   = . $\frac{2}{10}$

$\frac{1}{R}$=  $\frac{2}{10}$

R= 5 Ω

NEW CURRENT=  $\frac{V}{I}$

                         = . 20/ 5

                        = 4Α