Two electric bulbs rated ${P_1}\,watt$ $V\, volts$ and ${P_2}\, watt$ $V\, volts$ are connected in parallel and $V\, volts$ are applied to it. The total power will be
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If resistances of bulbs are ${R_1}$ and ${R_2}$ respectively then in parallel
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Power dissipated across the $8 \,\,\Omega$ resistor in the circuit shown here is $2\,\, watt.$ The power dissipated in watt units across the $3 \,\,\Omega$ resistor is
A potentiometer $PQ$ is set up to compare two resistances as shown in the figure. The ameter $A$ in the circuit reads $1.0\, A$ when two way key $K_3$ is open. The balance point is at a length $l_1\, cm$ from $P$ when two way key $K_3$ is plugged in between $2$ and $1$ , while the balance point is at a length $l_2\, cm$ from $P$ when key $K_3$ is plugged in between $3$ and $1$ . The ratio of two resistances $\frac{{{R_1}}}{{{R_2}}}$ is found to be
Two resistances of $400$ $\Omega$ and $800$ $\Omega$ are connected in series with $6\, volt$ battery of negligible internal resistance. A voltmeter of resistance $10,000$ $\Omega$ is used to measure the potential difference across $400$ $\Omega$. The error in the measurement of potential difference in volts approximately is
There are $n$ similar conductors each of resistance $R$. The resultant resistance comes out to be $x$ when connected in parallel. If they are connected in series, the resistance comes out to be
There are $0.8 \times 10^{23}$ free electrons $/ cm^3$ in copper. If $0.2\, A $ current is flowing is copper wire, then the drift velocity of electrons will be, if the cross sectional area of wire is $0.01 \,cm^2$