Find the electric potential at points C and D | vedclass

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

Question

(B) The point $P$ is grounded,so its potential $V_P = 0\,V$.The total electromotive force $(EMF)$ in the circuit is $E_{net} = 12\,V - 6\,V = 6\,V$.Th

Vedclass · Accepted Answer

$V_C = 9\,V$; $V_D = 6\,V$

Vedclass · Answer

(B) The point $P$ is grounded,so its potential $V_P = 0\,V$.The total electromotive force $(EMF)$ in the circuit is $E_{net} = 12\,V - 6\,V = 6\,V$.The total resistance in the circuit is $R_{eq} = 1\,\Omega + 2\,\Omega + 3\,\Omega = 6\,\Omega$.The current $I$ flowing in the circuit is $I = \frac{E_{net}}{R_{eq}} = \frac{6\,V}{6\,\Omega} = 1\,A$ (flowing clockwise).Starting from point $P$ $(0\,V)$ and moving towards $C$ through the $12\,V$ battery and $1\,\Omega$ and $2\,\Omega$ resistors:$V_C = V_P + 12\,V - I(1\,\Omega + 2\,\Omega) = 0 + 12 - 1(3) = 9\,V$.Starting from point $P$ $(0\,V)$ and moving towards $D$ through the $6\,V$ battery (in reverse direction) and $3\,\Omega$ resistor:$V_D = V_P - 6\,V + I(3\,\Omega) = 0 - 6 + 1(3) = -3\,V$. Wait,re-evaluating the circuit: The current flows from the $12\,V$ source through the resistors to the $6\,V$ source. $V_C = 12 - 1(1+2) = 9\,V$.$V_D = V_C - I(3) = 9 - 1(3) = 6\,V$.Thus,$V_C = 9\,V$ and $V_D = 6\,V$.

Find the electric potential at points $C$ and $D$ in the given circuit.

Similar Questions

$A$ part of a circuit is shown in the figure. All the ammeters are ideal. If the reading of ammeter $A_1$ is $1.0 \ A$,then:

If the galvanometer $G$ does not show any deflection in the circuit shown,the value of $R$ is given by $............\Omega$.

Inside a superconducting ring,six identical resistors each of resistance $R$ are connected as shown in the figure. The equivalent resistance$(s)$:

In the circuit shown,the resistance $r$ is a variable resistance. If for $r = fR$,the heat generation in $r$ is maximum,then the value of $f$ is:

Find the current passing through the battery immediately after the key $(K)$ is closed. It is given that initially all the capacitors are uncharged. (Given that $E = 5 \, V$, $R = 6 \, \Omega$) (in $A$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module