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
Medium
Download our appand 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.*
Similar Questions
- 1For which of the following dependence of drift velocity $v$ on electric field $E$ is ohm's law obeyedView Solution
- 2The equivalent resistance between $A$ and $B$ as shown in figure is $........\,k\Omega$View Solution
- 3In the circuit shown the cells are ideal and of equal emfs, the capacitance of the capacitor is $C$ and the resistance of the resistor is $R. X$ is first joined to $Y$ and then to $Z$. After a long time, the total heat produced in the resistor will beView Solution
- 4Ten $60 \,W , 220 \,V$ bulbs are connected in series to $220 \,V$ supply. Power consumed in the circuit is ......... $W$View Solution
- 5A battery of $24\,cells$ , each of emf $1.5\,V$ and internal resistance $2\,\Omega$ is to be connected in order to send the maximum current through a $12\,\Omega$ resistor. The correct arrangement of cells will beView Solution
- 6If three bulbs $60\,W$, $100\,W$ and $200\,W$ are connected in parallel, thenView Solution
- 7An electric lamp is marked $60\, W$, $230\, V$. The cost of a $1\, kWh$ of energy is Rs. $1.25$. The cost of using this lamp $8$ hrs a day for $30$ day is Rs. ...............View Solution
- 8View SolutionIf an observer is moving with respect to a stationary electron, then he observes
- 9A new flashlight cell of $e.m.f.$ $1.5\, volts$ gives a current of $15\, amps$, when connected directly to an ammeter of resistance $0.04\,\Omega $. The internal resistance of cell is ........... $\Omega$View Solution
- 10The figure shows a network of currents. The magnitude of currents is shown here. The current $I$ will be ........... $A$View Solution
