Find out current in $3\,\Omega $  resistance in given circuit
Medium
Download our app for free and get startedPlay store
$\mathrm{R}_{\mathrm{eq}}=[(4+2) \| 3]+5$

$=7\, \Omega$

$I=\frac{14}{7}=2 \,A$

$I_{1}=\frac{6}{9} \times 2=\frac{4}{3} \,A$

art

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.*

Similar Questions

  • 1
    A $25\, watt$, $220\, volt$ bulb and a $100\, watt$, $220\, volt$ bulb are connected in series across a $220\, volt$ lines. Which electric bulb will glow more brightly
    View Solution
  • 2
    Calculate equivalent resistance across $A-B$ ............... $\Omega$
    View Solution
  • 3
    Two batteries $A$ and $B$ each of $e.m.f.$ $2\, V$ are connected in series to an external resistance $R = 1 \,ohm$. If the internal resistance of battery $A$ is $1.9\, ohms$ and that of $B$ is $0.9\, ohm$, what is the potential difference between the terminals of battery $A$ ............. $V$
    View Solution
  • 4
    Assertion : When current through a bulb decreases by $0.5\%$, the glow of bulb decreases by $1\%$.

    Reason : Glow (Power) which is directly proportional to square of current.

    View Solution
  • 5
    The equivalent resistance between $P$ and $Q$ in the given figure, is ............... $\Omega$
    View Solution
  • 6
    Two ideal batteries of emf $V _1$ and $V _2$ and three resistances $R _1, R _2$ and $R _3$ are connected as shown in the figure.

    The current in resistance $R _2$ would be zero if

    $(A)$ $V_1=V_2$ and $R_1=R_2=R_3$

    $(B)$ $V_1=V_2$ and $R_1=2 R_2=R_3$

    $(C)$ $V_1=2 V_2$ and $2 R_1=2 R_2=R_3$

    $(D)$ $2 V _1= V _2$ and $2 R _1= R _2= R _3$

    View Solution
  • 7
    In the circuit shown in figure, all cells are ideal. The current through $2 \,\Omega$ resistor is ............ $A$
    View Solution
  • 8
    If current in an electric bulb changes by $1\%$, then the power will change by ............ $\%$
    View Solution
  • 9
    A group of $N$ cells whose $emf$ varies directly with the internal resistance as per the equation $E_N = 1.5\, r_N$ are connected as shown in the figure below. The current $I$ in the circuit is ........... $amp$
    View Solution
  • 10
    In a thin rectangular metallic strip a constant current $I$ flows along the positive $x$-direction, as shown in the figure. The length, width and thickness of the strip are $\ell$, w and $d$, respectively. A uniform magnetic field $\vec{B}$ is applied on the strip along the positive $y$-direction. Due to this, the charge carriers experience a net deflection along the $z$ direction. This results in accumulation of charge carriers on the surface $P Q R S$ and appearance of equal and opposite charges on the face opposite to $PQRS$. A potential difference along the $z$-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons.

    $1.$ Consider two different metallic strips ($1$ and $2$) of the same material. Their lengths are the same, widths are $w_1$ and $w_2$ and thicknesses are $d_1$ and $d_2$, respectively. Two points $K$ and $M$ are symmetrically located on the opposite faces parallel to the $x$ - $y$ plane (see figure). $V _1$ and $V _2$ are the potential differences between $K$ and $M$ in strips $1$ and $2$ , respectively. Then, for a given current $I$ flowing through them in a given magnetic field strength $B$, the correct statement$(s)$ is(are)

    $(A)$ If $w _1= w _2$ and $d _1=2 d _2$, then $V _2=2 V _1$

    $(B)$ If $w_1=w_2$ and $d_1=2 d_2$, then $V_2=V_1$

    $(C)$ If $w _1=2 w _2$ and $d _1= d _2$, then $V _2=2 V _1$

    $(D)$ If $w _1=2 w _2$ and $d _1= d _2$, then $V _2= V _1$

    $2.$ Consider two different metallic strips ($1$ and $2$) of same dimensions (lengths $\ell$, width w and thickness $d$ ) with carrier densities $n_1$ and $n_2$, respectively. Strip $1$ is placed in magnetic field $B_1$ and strip $2$ is placed in magnetic field $B_2$, both along positive $y$-directions. Then $V_1$ and $V_2$ are the potential differences developed between $K$ and $M$ in strips $1$ and $2$, respectively. Assuming that the current $I$ is the same for both the strips, the correct option$(s)$ is(are)

    $(A)$ If $B_1=B_2$ and $n_1=2 n_2$, then $V_2=2 V_1$

    $(B)$ If $B_1=B_2$ and $n_1=2 n_2$, then $V_2=V_1$

    $(C)$ If $B _1=2 B _2$ and $n _1= n _2$, then $V _2=0.5 V _1$

    $(D)$ If $B_1=2 B_2$ and $n_1=n_2$, then $V_2=V_1$

    Give the answer question $1$ and $2.$ 

    View Solution