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six lead -acid type of secondary cells ,each of emf 2v and internal resistance 0.015 are joined in series to provide supply to a resistance of 8.5.Determine (i) the current drawn from the supply and (ii) its terminal voltage

answer=1.4A,11.9V

explain with step

i am not getting th e answer

Deepak Patra , 12 Years ago
Grade upto college level
anser 1 Answers
Askiitians Tutor Team

To solve this problem, we need to analyze the circuit involving the six lead-acid batteries connected in series and the external resistance. Let's break it down step by step.

Understanding the Components

We have six lead-acid batteries, each with the following specifications:

  • Electromotive force (emf) = 2 V
  • Internal resistance = 0.015 Ω

When batteries are connected in series, their voltages add up, while the internal resistances also add up. The external resistance connected to the batteries is 8.5 Ω.

Calculating Total Voltage and Resistance

First, let's find the total voltage provided by the six batteries:

  • Total emf = Number of batteries × emf of each battery
  • Total emf = 6 × 2 V = 12 V

Next, we calculate the total internal resistance of the six batteries:

  • Total internal resistance = Number of batteries × internal resistance of each battery
  • Total internal resistance = 6 × 0.015 Ω = 0.09 Ω

Applying Ohm's Law

Now, we can find the total resistance in the circuit. This includes both the total internal resistance of the batteries and the external resistance:

  • Total resistance (R_total) = Total internal resistance + External resistance
  • R_total = 0.09 Ω + 8.5 Ω = 8.59 Ω

Finding the Current Drawn from the Supply

Using Ohm's Law (V = IR), we can find the current (I) drawn from the supply:

  • Current (I) = Total voltage / Total resistance
  • I = 12 V / 8.59 Ω ≈ 1.4 A

Calculating the Terminal Voltage

Now that we have the current, we can determine the terminal voltage of the battery supply. The terminal voltage (V_terminal) can be calculated using the formula:

  • V_terminal = Total emf - (Current × Total internal resistance)
  • V_terminal = 12 V - (1.4 A × 0.09 Ω)
  • V_terminal = 12 V - 0.126 V ≈ 11.874 V
  • Rounding this gives us approximately 11.9 V.

Summary of Results

In summary, the current drawn from the supply is approximately 1.4 A, and the terminal voltage is about 11.9 V. This analysis shows how to combine the effects of series connections in both voltage and resistance to find the current and terminal voltage in a circuit.

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