Find the current in the 10,Omega resistance. | vedclass

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

Question

(C) Let the potential at the bottom node be $0\,V$ and the potential at the top node be $V$. Using Kirchhoff's Current Law $(KCL)$ at the top node:$\f

Vedclass · Accepted Answer

$0.45\, A$ from $P_2$ to $P_1$

Vedclass · Answer

(C) Let the potential at the bottom node be $0\,V$ and the potential at the top node be $V$. Using Kirchhoff's Current Law $(KCL)$ at the top node:$\frac{V - 5}{2} + \frac{V}{10} + \frac{V - 5}{2} = 0$Multiply by $10$ to simplify:$5(V - 5) + V + 5(V - 5) = 0$$5V - 25 + V + 5V - 25 = 0$$11V = 50$$V = \frac{50}{11} \approx 4.545\,V$The current $I$ through the $10\,\Omega$ resistor flows from the top node $(P_2)$ to the bottom node $(P_1)$:$I = \frac{V}{10} = \frac{50/11}{10} = \frac{5}{11} \approx 0.4545\,A$Rounding to two decimal places,the current is $0.45\,A$ from $P_2$ to $P_1$.

Find the current in the $10\,\Omega$ resistance. The circuit consists of two $5\,V$ batteries with internal resistances of $2\,\Omega$ each,connected in parallel with a $10\,\Omega$ resistor.

Similar Questions

As shown in the circuit,if a conducting wire is connected between points $A$ and $B$,then the electric current in this wire will flow:

$AB$ is a wire of uniform resistance. The galvanometer $G$ shows no current when the length $AC = 20\,cm$ and $CB = 80\,cm$. The resistance $R$ is equal to .............. $\Omega $

The value of current through the $20\,\Omega$ resistor is ............ $A$.

In the given circuit,the current through $AB$ is zero. What will be the value of the unknown resistance $X$ in $\Omega$?

The resistances in the Wheatstone bridge circuit shown in the diagram are all different,and the current through the galvanometer is zero. If all thermal effects are negligible,the current through the galvanometer will not be zero when:

Vedclass Test Series

Exam Paper Generator

Online Exam Module