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How does the sign of the phase angle phi...

How does the sign of the phase angle `phi`,by the supply voltage leads the current in an LCR series circuit,change as the supply frequency is gradually increased from very low to very high values?

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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 1.02V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire.To ensure that very low current 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 E 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 m V (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 1.02V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire.To ensure that very low current 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 E 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 k Omega 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 1.02V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire.To ensure that very low current 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 E and the balance point found similarly,turns out to be at 82.3 cm length of the wire . What is the value of E ?

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 1.02V (for very moderate currents upto a few mA) gives a balance point at 67.3 cm length of the wire.To ensure that very low current 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 E 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 ?

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