A torch bulb rated as $4.5\, W$, $1.5\, V$ is connected as shown in the figure. The $e.m.f.$ of the cell needed to make the bulb glow at full intensity is ................ $V$
A$4.5$
B$1.5$
C$2.67$
D$13.5$
Diffcult
Download our app for free and get started
D$13.5$
d (d) Current in the bulb $ = \frac{P}{V} = \frac{{4.5}}{{1.5}} = 3\,A$
Current in $1$ $\Omega$ resistance $ = \frac{{1.5}}{1} = 1.5\,A$
Hence total current from the cell $i = 3 + 1.5 = 4.5\,A$
By using $E = V + ir$==> $E = 1.5 + 4.5 \times (2.67) = 13.5\,V$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A battery is connected to a uniform resistance wire $AB$ and $B$ is earthed. Which one of the graphs below shows how the current density $J$ varies along $AB$
Two wires of the same material and equal length are joined in parallel combination. If one of them has half the thickness of the other and the thinner wire has a resistance of $8\, ohms$, the resistance of the combination is equal to
Equal potentials are applied on an iron and copper wire of same length. In order to have the same current flow in the two wires, the ratio $r$ (iron)/$r$ (copper) of their radii must be (Given that specific resistance of iron = $1.0 \times {10^{ - 7}}$ $ ohm-m$ and specific resistance of copper = $1.7 \times {10^{ - 8}}\,ohm-m$)
A voltmeter having a resistance of $998\, ohms$ is connected to a cell of $e.m.f.$ $2\, volt$ and internal resistance $2\, ohm$. The error in the measurement of $e.m.f.$ will be
The voltage of clouds is $4 \times 10^6\,volt$ with respect to round. In a lighteing strike lasting $100\,m\,sec$, a charge of $4\,coulombs$ is delivered to the ground. The power of lightening strike is
A wire of resistance $10$ $\Omega$ is bent to form a circle. $P$ and $Q$ are points on the circumference of the circle dividing it into a quadrant and are connected to a Battery of $3\, V$ and internal resistance $1$ $\Omega$ as shown in the figure. The currents in the two parts of the circle are