A short circuit occurs in a telephone cable having a resistance of` 0.45 Omega m^(-1). `The circuit is tested with a Wheat stone bridge network have values of `100 Omega and 1,110 Omega` respectively. A balance condition is found when the variable resistor has a value of` 400 Omega.` Calculate the distance down the cable, where the short has occured.

The potentiometer wire of length 100cm has a resistance of 10Omega. It is connected in series with a resistance of 5Omega and an accumulat or emf 3V having negligible resistance. A source 1.2V is balanced against a length 'L' of the potentionmeter wire. find the value of L.

A wire having a mass of 0.45 kg posseses a resistanfce of 0.014 Omega . If the resistivity of the material fo wire is 1.78 xx10^-7 Omegam , calculate its length and radius. Given that the density of the material of wire is 8.93 xx10^3 kg m^-3

Two cells, E_(1) and E_(2) of emfs 4 V and 8 V having internal resistances 0.5 Omega and 1.0 Omega respectively are connected in opposition to each other. This combination is connected in opposition to with resistance of 4.5 Omega and 3.0 Omega . Another resistance of 6 Omega is connected in parallel across the 3 Omega resistor. Draw the circuit diagram

A galvanometer having a coil resistance of 100 omega gives a full scale deflection , when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10A.

Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k Omega is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsilon and the balance point found similarly, turns out to be at 82.3 cm length of the wire. Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?:

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k Omega is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsilon and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What purpose does the high resistance of 600 kOmega have? :

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k Omega is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsilon and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What is the value epsilon ? :

Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k Omega is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsilon and the balance point found similarly, turns out to be at 82.3 cm length of the wire. Is the balance point affected by the internal resistance of the driver cell?:

Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 k Omega is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf epsilon and the balance point found similarly, turns out to be at 82.3 cm length of the wire. Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V? :