A series combination of two resistors 1,Omega | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The total resistance of the circuit in series is $R_{eq} = R_1 + R_2 = 1\,\Omega + 1\,\Omega = 2\,\Omega$.The total resistance including the inter

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The total resistance of the circuit in series is $R_{eq} = R_1 + R_2 = 1\,\Omega + 1\,\Omega = 2\,\Omega$.The total resistance including the internal resistance $r$ of the battery is $R_{total} = R_{eq} + r = 2\,\Omega + 0.4\,\Omega = 2.4\,\Omega$.Using Ohm's law,the current $I$ flowing through the circuit is given by $I = \frac{V}{R_{total}}$.Substituting the values,$I = \frac{12\,V}{2.4\,\Omega} = 5\,A$.Therefore,the correct option is $B$.

$A$ series combination of two resistors $1\,\Omega$ each is connected to a $12\,V$ battery of internal resistance $0.4\,\Omega$. The current flowing through the circuit will be ............... $A$.

Similar Questions

If two bulbs of wattage $25\, W$ and $100\, W$ respectively,each rated at $220\, V$,are connected in series with a supply of $440\, V$,then which bulb will fuse?

For the circuits $A$ and $B$ as shown in the figure,identify the correct option.

The equivalent resistance between $A$ and $B$ is ............ $\Omega$.

In the given circuit,the current flowing through the circuit is: (in $A$)

$A$ voltmeter of resistance $998\ \Omega$ is connected to a cell of $emf$ $2\ V$ and internal resistance $2\ \Omega$. The error in the measurement of $emf$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module