Sure, let's solve this problem step-by-step.
We have two identical cells each with an electromotive force (emf) of [tex]\(1.5 \text{ V}\)[/tex] and an internal resistance of [tex]\(0.5 \Omega\)[/tex]. These cells are connected in series to a coil with a resistance of [tex]\(5 \Omega\)[/tex]. We need to determine:
a) The effective emf of the cells in series.
b) The effective resistance in the circuit.
c) The current in the circuit.
d) The voltage drop across the solution resistor (coil).
### a) Effective EMF of the cells in series:
When cells are connected in series, their emf adds up. Therefore, the total emf ([tex]\(E_{total}\)[/tex]) is:
[tex]\[ E_{total} = 1.5 \text{ V} + 1.5 \text{ V} = 3.0 \text{ V} \][/tex]
### b) Effective resistance in the circuit:
The total internal resistance ([tex]\(r_{total}\)[/tex]) of the two cells in series is the sum of their individual resistances:
[tex]\[ r_{total} = 0.5 \Omega + 0.5 \Omega = 1.0 \Omega \][/tex]
Now, the total resistance ([tex]\(R_{total}\)[/tex]) in the circuit is the sum of the total internal resistance of the cells and the resistance of the coil:
[tex]\[ R_{total} = r_{total} + \text{resistance of the coil} \][/tex]
[tex]\[ R_{total} = 1.0 \Omega + 5 \Omega = 6.0 \Omega \][/tex]
### c) Current in the circuit:
Using Ohm's Law, the current ([tex]\(I\)[/tex]) in the circuit is given by:
[tex]\[ I = \frac{E_{total}}{R_{total}} \][/tex]
[tex]\[ I = \frac{3.0 \text{ V}}{6.0 \Omega} = 0.5 \text{ A} \][/tex]
### d) Voltage drop across the coil:
The voltage drop across the coil ([tex]\(V_{coil}\)[/tex]) can be found using Ohm's Law again:
[tex]\[ V_{coil} = I \times \text{resistance of the coil} \][/tex]
[tex]\[ V_{coil} = 0.5 \text{ A} \times 5 \Omega = 2.5 \text{ V} \][/tex]
### Summary:
a) Effective emf of the cells in series: [tex]\(3.0 \text{ V}\)[/tex]
b) Effective resistance in the circuit: [tex]\(6.0 \Omega\)[/tex]
c) Current in the circuit: [tex]\(0.5 \text{ A}\)[/tex]
d) Voltage drop across the coil: [tex]\(2.5 \text{ V}\)[/tex]
