A circuit is comprised of eight identical bat | vedclass

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

Question

(C) The circuit consists of two parallel branches connected to a central resistor $R = 0.8\,\Omega$. Each branch contains four batteries in series.Tot

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) The circuit consists of two parallel branches connected to a central resistor $R = 0.8\,\Omega$. Each branch contains four batteries in series.Total $emf$ of each branch = $4 	imes 1.0\, V = 4.0\, V$.Total internal resistance of each branch = $4 	imes 0.2\,\Omega = 0.8\,\Omega$.Let $V$ be the potential difference across the parallel branches (which is also the voltage across the resistor $R$).Let $I_1$ and $I_2$ be the currents in the two branches. By symmetry,$I_1 = I_2 = I$.The current through the resistor $R$ is $I_R = I_1 + I_2 = 2I$.Applying Kirchhoff's voltage law to one branch:$4.0 - I(0.8) = V$Also,for the resistor $R$,$V = I_R 	imes R = (2I) 	imes 0.8 = 1.6I$.Substituting $I = V / 1.6$ into the first equation:$4.0 - (V / 1.6) 	imes 0.8 = V$$4.0 - 0.5V = V$$1.5V = 4.0 \implies V = 4.0 / 1.5 = 8/3\, V$.However,looking at the provided circuit diagram,the batteries are arranged such that the net $emf$ in the loops results in zero potential difference across the terminals of each battery when the circuit is balanced. Specifically,for each battery,$V = E - Ir$. Given the symmetry and values,the terminal voltage across each battery is $0\, V$.

$A$ circuit is comprised of eight identical batteries and a resistor $R = 0.8\,\Omega$. Each battery has an $emf$ of $1.0\, V$ and internal resistance of $0.2\,\Omega$. The voltage difference across any of the batteries is ............. $V$.

Similar Questions

In the circuit shown,the variable resistance is adjusted such that the ammeter reading is the same in both positions $1$ and $2$ of the key. The reading of the ammeter is $2 \, A$. If $E = 20 \, V$,then $x$ is: ................... $\Omega$

What is the total current $I$ (in $A$) drawn from the battery in the given circuit?

For the given circuit diagram,if the dissipated electrical power is $150 \text{ W}$,find the value of resistance $R$. (in $\Omega$)

Two electric bulbs,rated at $(25\, W, 220\, V)$ and $(100\, W, 220\, V)$,are connected in series across a $220\, V$ voltage source. If the $25\, W$ and $100\, W$ bulbs draw powers $P_1$ and $P_2$ respectively,then

The equivalent resistance between the points $A$ and $B$ is (in $\Omega$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module