A potentiometer circuit consists of a battery of e.m.f 5v and internal resistance 10 ohms connected in series with a 30 ohms resistor and a potentiometer wire AB of length 1m and resistance 20 ohms calculate:the total resistance of the circiuit ,the current flowing in the cicuit,the p.d across the wire AB.?
A potentiometer circuit consists of a battery of emf 5v and internal resistance 1.0 ohm connected in series with a potentiometer wire AB of length 1.0m and resistance 2.0 ohms. calculate the emf of a dry cell which can be balanced across 60cm of the wire AB
To calculate the total resistance of the circuit, we need to add up all the resistances in series. The circuit consists of the internal resistance of the battery (10 ohms), the resistor (30 ohms), and the potentiometer wire AB (20 ohms). Therefore, the total resistance of the circuit is:
Total resistance = 10 ohms + 30 ohms + 20 ohms
= 60 ohms
Next, we can calculate the current flowing in the circuit using Ohm's Law (I = V / R), where V is the voltage and R is the resistance. In this case, the voltage is given as 5V and the total resistance is 60 ohms. Hence, the current flowing in the circuit is:
Current (I) = Voltage (V) / Resistance (R)
= 5V / 60 ohms
= 0.0833 A or 83.3 mA (rounded to three decimal places)
Finally, to find the potential difference across the wire AB, we'll use the concept of voltage division. The voltage across the wire AB can be calculated as the fraction of the total voltage (5V) determined by the resistance of the wire AB relative to the total resistance. The resistance of the wire AB is given as 20 ohms, and the total resistance of the circuit is 60 ohms. Therefore, the potential difference across the wire AB is:
P.D. across AB = (Resistance of AB / Total Resistance) * Voltage
= (20 ohms / 60 ohms) * 5V
= (1/3) * 5V
= 5/3 V or 1.67 V (rounded to two decimal places)
total resistance=10+30+20= ?
current= 5v/totalresistance
I dont know where ab is, if it is the battery, then Vab=5-10*current