Question

two resistors with resistance 5 ohm and 10 ohm respectively are to be connected to a battery of 6 volt so as to abtain (i) minimum current (ii) maximum current. (1) how will you connect the resistances in each case (2) calculate the strength of the total current in the circuit in two cases​

Answer 1

Answer :

We are provided two resistors with resistance 5Ω and 10Ω.

Voltage of battery = 6V

For maximum current current flow in the circuit, we have to connect them in parallel.

For minimum current flow in the circuit, we have to connect them in series$.$

We can solve this question by using ohm's law. As per this law, current flow through a circuit is directly proportional to the applied potential difference$.$ [V = IR]

◈ Current flow in series :

➝ V = I × Rs

➝ V = I × (R¹ + R²)

➝ 6 = I × (5 + 10)

➝ I = 6/15

➝ I = 0.4A

◈ Current flow in parallel :

➝ V = I × Rp

➝ V = I × (R¹R²/R¹+R²)

➝ 6 = I × (5×10/5+10)

➝ 6 = I × (50/15)

➝ I = (6×15/50)

➝ I = 1.8A

Answer 2

Answer:

i) minimum current =0.4 A

ii) maximum current=6/3.3= 1.9 approx

1)in first case resistance are in series to obtain minimum current and in second case resistance are in parallel to obtain maximum current.

2) in first case total current is 0.4 ampere

in second case total current is 1.9 ampere

Explanation:

according to the current electricity chapter

we know that ,if the resistance are combined in series then it give the minimum of current in a circuit.

while if the resistance are combined in the parallel then it will give maximum current in circuit.

so according to the ohm's law we can find the current in the circuit by the given formula

R= V/ I

In first case

total resistance in circuit R=R1+R2

R= 5+10

R=15 ohm

so I = V/ R

l= 6/15

I=0.4 ampere