Given data:
- Battery voltage, V = 10 V
- Internal resistance of battery, r = 1 ohm
- Characteristics of load: 7i = v^2 + 2v

To find: Current delivered by battery

Solution:

1. Find the equivalent resistance of the load:
- The load characteristics equation can be written as: v^2 + 2v - 7i = 0
- This is a quadratic equation, and its solution is: v = [-2 ± sqrt(4 + 28i)]/2
- Since the load is non-linear, its resistance is not constant. However, we can find its equivalent resistance for a given current i as:
R_eq = (v1 - v2)/i, where v1 and v2 are the two possible values of v for a given i
- Substituting the value of v in the above equation, we get: R_eq = (sqrt(4 + 28i) - (-2 - sqrt(4 + 28i)))/(14i/7)
- Simplifying this expression, we get: R_eq = 2/(7i)

2. Find the total circuit resistance:
- The total circuit resistance is the sum of the load resistance and the internal resistance of the battery, i.e., R_total = R_eq + r
- Substituting the value of R_eq and r, we get: R_total = 2/(7i) + 1

3. Find the current delivered by the battery:
- The current delivered by the battery is given by: I = V/R_total
- Substituting the value of V and R_total, we get: I = 10/(2/(7i) + 1)
- Simplifying this expression, we get: I = 70i/(7i + 20)

Therefore, the current delivered by the battery is 70i/(7i + 20).

Explanation:

- The problem involves finding the current delivered by a battery connected across a non-linear load whose characteristics are given.
- The load characteristics equation is a quadratic equation, which can be solved to find the voltage across the load for a given current.
- The equivalent resistance of the load can then be found using Ohm's law, and the total circuit resistance can be calculated as the sum of the load resistance and the internal resistance of the battery.
- Finally, the current delivered by the battery can be found using Ohm's law again.

The relevant circuit is shown in Fig.

Non-linear load characteristics:
7I = V2 + 2V ……….. (i)

Loop equation:
I + V = 10 ……….. (ii)

Solving (i) and (ii)
V2 + 2V = 7 (10 – V)
V2 + 2V = 70 – 7V
V2 + 9V – 70 = 0
(V – 5) (V + 14) = 0
V = 5V, –14V

As V cannot be negative, V is taken as 5V
∴ I = 10 – V = 10 – 5 = 5A