The electric current passing through a metallic wire produces heat because of
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(d) Colliding electrons lose their kinetic energy as heat.
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A potentiometer having the potential gradient of $2\, mV/cm$ is used to measure the difference of potential across a resistance of $10 \,\Omega$. If a length of $50\, cm$ of the potentiometer wire is required to get the null point, the current passing through the $10 \,\Omega$ resistor is (in $mA$)
In a potentiometer arrangement, a cell of emf $1.20\, V$ gives a balance point at $36\, cm$ length of wire. This cell is now replaced by another cell of emf $1.80\, V$. The difference in balancing length of potentiometer wire in above conditions will be $....cm$.
A milliammeter of range $10\, mA$ and resistance $9\, \Omega$ is joined in a circuit as shown. The meter gives full-scale deflection for current $I$ when $A$ and $B$ are used as its terminals, i.e., current enters at $A$ and leaves at $B$ ($C$ is left isolated). The value of $I$ is
An electric iron draws $5\, amp$, a $TV$ set draws $3\, amp$ and refrigerator draws $2\, amp$ from a $220\, volt$ main line. The three appliances are connected in parallel. If all the three are operating at the same time, the fuse used may be of ............. $amp$
Water of volume $2\, litre$ in a container is heated with a coil of $1\, kW$ at $27 \,^oC$. The lid of the container is open and energy dissipates at rate of $160\, J/s$. In how much time temperature will rise from $27\,^oC$ to $77\,^oC$ $[$ Given specific heat of water is $4.2\, kJ/kg$ $]$
One $kg$ of water, at $20\,^oC$, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of $20\, \Omega $. The rms voltage in the mains is $200\, V$. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to.......... $\min$ [Specific heat of water $= 4200\, J/kg\, ^oC$), Latent heat of water $= 2260\, k\,J/kg$]
Under what conditions current passing through a resistance $R$ can be increased by short circuiting the battery of emf $E_2$. The internal resistances of the two batteries are $r_1$ and $r_2$ respectively
In a typical Wheatstone network, the resistances in cyclic order are $A = 10 \,\Omega $, $B = 5 \,\Omega $, $C = 4 \,\Omega $ and $D = 4 \,\Omega $ for the bridge to be balanced
If you are provided three resistances $2 \,\Omega$, $3 \,\Omega$ and $6 \,\Omega$. How will you connect them so as to obtain the equivalent resistance of $4 \,\Omega$