Use Kirchhoff's laws, calculate the potential difference across the $8\Omega$ Resistor in the circuit of Fig.$5.161.$

 

A battery $E_1$ of $4V$ And a variable resistance $Rh$ Are connected in series with the wire $AB$ Of the potentiometer Fig.$5.162.$ The length of the wire of the potentiometer is $1$ Metre. When a cell $E_2$ Of emf $1.5V$ Is connected between $A$ and $C,$ No current flows through $E_2$, Length of $AC=60 cm,$

(i) Find the potential difference between the ends $A$ and $B$ Of the potentiometer,

A battery $E_1$ of $4V$ And a variable resistance $Rh$ Are connected in series with the wire $AB$ Of the potentiometer Fig.5.162.5.162. The length of the wire of the potentiometer is 11 metre. When a cell $E_2$ Of emf $1.5V$ Is connected between $A$ and $C,$ No current flows through $E_2$, Length of $AC=60 cm,$

(ii) would the method work if the battery $E_1$ Is replaced by a cell of emf $1V$ ?

The figure, Fig. $5.163$ Shows a potentiometer with a cell of $2.0 \mathrm{V}$ And internal resistance $0.40\Omega$ Maintaining a potential.

Drop across a resistor wire $\mathrm{AB}$ A standard cell which maintains a constant emf of $1.02 \mathrm{V}$ (for very moderate current currents up to a few $\mu \mathrm{A}$ ) gives a balance point at $67.3 \mathrm{cm}$ Length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of $600\mathrm{k}\Omega$ Is put in series with it, which is shorted close to balance point. The standard cell is then replaced by a cell of unknown emf $\in$ And the balance point found similarly turns out to be at $82.3 \mathrm{cm}$ Length of the wire.

(a) What is the value of $\in$?

The figure, Fig. $5.163$ Shows a potentiometer with a cell of $2.0 \mathrm{V}$ And internal resistance $0.40\Omega$ Maintaining a potential

Drop across a resistor wire $\mathrm{AB}$ A standard cell which maintains a constant emf of $1.02 \mathrm{V}$ (for very moderate current currents up to a few $\mu \mathrm{A}$ ) gives a balance point at $67.3 \mathrm{cm}$ Length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of $600\mathrm{k}\Omega$ Is put in series with it, which is shorted close to balance point. The standard cell is then replaced by a cell of unknown emf $\in$ And the balance point found similarly turns out to be at $82.3 \mathrm{cm}$ Length of the wire.

(b) What purpose does the high resistance of $6000\mathrm{k}\Omega$.

The figure, Fig. $5.163$ Shows a potentiometer with a cell of $2.0 \mathrm{V}$ And internal resistance $0.40\Omega$ Maintaining a potential

Drop across a resistor wire $\mathrm{AB}$ A standard cell which maintains a constant emf of $1.02 \mathrm{V}$ (for very moderate current currents up to a few $\mu \mathrm{A}$ ) gives a balance point at $67.3 \mathrm{cm}$ Length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of $600\mathrm{k}\Omega$ Is put in series with it, which is shorted close to balance point. The standard cell is then replaced by a cell of unknown emf $\in$ And the balance point found similarly turns out to be at $82.3 \mathrm{cm}$ Length of the wire.

(c) Is the balance point effect by this high resistance?

The figure, Fig. $5.163$ Shows a potentiometer with a cell of $2.0 \mathrm{V}$ And internal resistance $0.40\Omega$ Maintaining a potential

Drop across a resistor wire $\mathrm{AB}$ A standard cell which maintains a constant emf of $1.02 \mathrm{V}$ (for very moderate current currents up to a few $\mu \mathrm{A}$ ) gives a balance point at $67.3 \mathrm{cm}$ Length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of $600\mathrm{k}\Omega$ Is put in series with it, which is shorted close to balance point. The standard cell is then replaced by a cell of unknown emf $\in$ And the balance point found similarly turns out to be at $82.3 \mathrm{cm}$ Length of the wire.

(d) Is the balance point affected by the internal resistance of the driver cell?

The figure, Fig. $5.163$ Shows a potentiometer with a cell of $2.0 \mathrm{V}$ And internal resistance $0.40\Omega$ Maintaining a potential

Drop across a resistor wire $\mathrm{AB}$ A standard cell which maintains a constant emf of $1.02 \mathrm{V}$ (for very moderate current currents up to a few $\mu \mathrm{A}$ ) gives a balance point at $67.3 \mathrm{cm}$ Length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of $600\mathrm{k}\Omega$ Is put in series with it, which is shorted close to balance point. The standard cell is then replaced by a cell of unknown emf $\in$ And the balance point found similarly turns out to be at $82.3 \mathrm{cm}$ Length of the wire.

(e) Would the method work in the above solution if the driver cell of potentiometer had an emf of $1.0 \mathrm{V}$ Instead of $2.0 \mathrm{V}$?