You buy a AA battery in the store, and it is marked 1.5 V. If this marking is strictly accurate, while this battery is fresh its terminal voltage will be more than 1.5 V when it is used in a circuit. its terminal voltage will be less than 1.5 V when it is used in a circuit. its terminal voltage will be 1.5 V when it is used in a circuit.

Option A is correct.

when it is used in a circuit. its terminal voltage will be less than 1.5 V.

Explanation:

The terminal voltage of the battery when it is in use in circuits drops lower than the 1.5 V rating given to it due to internal resistance.

All batteries give internal resistances when used in circuits. The internal resistance (though very small) is usually modelled as connected in series with the battery. It is due to some form of interference from the chemical makeup of the battery.

Normally, while the battery is fresh, the voltage (V) obtained at its terminals when connected in series with a resistor of resistance R is V = IR; where I is the current flowing in this circuit.

But once the interenal resistance (r) of the battery comes into play,

V = I₁ (r + R)

The current in the circuit evidently drops (that is I₁ < I) and V = (I₁r + I₁R)

The voltage across the terminals of the battery is no longer V but is now (V) × [R/(R+r)] which is less than the initial V and it reduces as the internal resistance, r, increases.

Hope this Helps!!!

the answer is the first electromagnet (a). this is because it has more coils around the ferromagnetic material. increasing the number of coils increases the surface area in which the metal interacts with the current in the coil. this enables more atoms in the metal to become aligned hence becoming more magnetized.