Calculate: \(I_{1}\), \(I_{2}\) and \(I_{3}\) in below diagram;
Ans: \(I_{1}=2A\), \(I_{2}=4A\) and \(I_{3}=-2A\).
Using Kirchhoff’s laws in the given electric network shown in below figure. Calculate: the value of \(I_{1}\), \(I_{2}\) and \(I_{3}\).
Ans: \(I_{1}=\frac{48}{31}A\). \(I_{2}=\frac{18}{31}A\) and \(I_{3}=\frac{66}{31}A\)
Use kirchhoff’s rule to determine the potential difference between the points A and D when no current flows through arm BE, of the electrical network shown bellow:
Ans: \(V_{CD}=9V\).
Find the potential difference across each cell and the rate of energy dissipation inR.
Ans: \(\frac{48}{7}V\) and \(11.75 W\).
A battery of \(6V\) and internal resistance \(0.5\Omega\) is joined in parallel with another of \(10V\) and internal resistance of \(1\Omega\). The combination sends a current through an external resistance of \(12\Omega\). Find: the current through each battery?
Ans: \(I_{1}=-\frac{84}{37}A\), \(I_{2}=\frac{106}{37}A\) and \(I=-\frac{22}{37}A\).
The four arms of a Wheatstone bridge have following resistances, \(AB=100\Omega\), \(BC=10\Omega\), \(CD=5\Omega\) and \(DA=60\Omega\). A galvanometer of \(15\Omega\) resistance is connected across BD. Calculate: the current through the galvanometer when a potential difference of \(10V\) is maintained across AC.
Ans: \(I_{g}=4.87mA\).
Two cells of e.m.f.’s \(1.5V\) and \(2V\) and internal resistances \(2\Omega\) and \(1\Omega\) respectively have their negative terminals joined by a wire of \(6\Omega\) and positive terminal by a wire of \(4\Omega\) resistance. A third resistance wire of \(8\Omega\) connects middle points of these wires. Draw the circuit diagram. Using Kirchhoff’s law find the potential difference at the end of this third wire.
Ans: \(V_{8\Omega}=1.26V\).
Calculate: the equivalent resistance between points A and B of the network shown:
Ans: \(R_{equ}=\frac{r\left(3R+r\right)}{R+3r}\).
Calculate: the current in the various branches of \(10\Omega\) resistance of the network of resistance as shown. Also, calculate: the total resistance between point A and B.
Ans: \(R_{eff}=7\Omega\).
Twelve wires each having a resistance of \(1\Omega\) are connected to form a cube. Find: the resistance of cube between two corners of the diagonal of one face of cube.
Ans: \(R_{equ}=\frac{3}{4}\Omega\).
Using Kirchhoff’s Rules. Calculate: the current through the \(40\Omega\) and \(20\Omega\) resistors in the following circuit.
Ans: \(I_{40\Omega}=0\) and \(I_{20\Omega}\).
Twelve wires each of having resistance of \(r\Omega\) are connected to form a skeleton cube. Find: the resistance of the cube between two corners of same edge.
Ans: \(R_{eff}=\frac{7}{12}r\).
Find: the equivalent resistance of the network shown in figure, between point A and B.
Ans: \(R_{equ}=r\).
Calculate: the current drawn from battery by the network of resistors shown in diagram below:
Ans: \(I=2A\)
Six equal resistance each of value R are joined together. Calculate: the equivalent resistance across AB. if the supply of e.m.f. is E is connected across AB, compute the current through the arms DE and AB.
Ans: \(R_{equ}=\frac{R}{2}\), and \(I_{AB}=\frac{2E}{R}\) & \(I_{DE}=0\).
A potential difference of \(10V\) is applied between the points A and B as shown in network below. Calculate: (i) equivalent resistance of the electrical network between points A and B. (ii) The current through the arms AFCEB and AFDEB. Value of R is assumed to be \(2\Omega\)
Ans:(i) \(R_{equ}=2\Omega\). (ii) \(I_{AFCEB}=I_{AFDEB}=2.5A\).
Find: the value of unknown resistance X, in he following circuit. If, no current flows through the section AO. Also; calculate the current drawn by the circuit from the battery of e.m.f. \(6V\) and negligible internal resistance.
Ans: \(I=1A\).
In the given network of resistances. Calculate: The value of X so that potential difference between B and D is zero.
Ans: \(X=18\Omega\).
The galvanometer, in each of the two given circuits does not show any deflection. Find: the ratio of the resistors \(R_{1}\) and \(R_{2}\) used in two circuits.
