1) resistance of circuit when key K is open
2) current drawn from cell when key is open
3) resistance of circuit when key k is closed
4) current drawn from cell when key is closed
Answers
1) resistance of circuit when key K is open
When the key is open, the only resistance left will be the 5 Ω resistance in series with 0.5 Ω source resistance,
hence net resistance = 5 + 0.5 = 5.5 Ω
2) current drawn from cell when key is open
When the key is open, net resistance = 5.5 Ω
=> current = voltage/ resistance
=> I = 3.3/5.5 = 0.6 A
hence current will be 0.6 Ampere.
3) resistance of circuit when key k is closed
When key is closed, the 5 Ω will be in parallel with series combination of 2 and 3Ω
hence net resistance = 0.5 + 5||(2+3)
= 0.5 + 5||5
= 0.5 + 2.5
= 3 Ω
4) current drawn from cell when key is closed
When the key is closed, net resistance = 3 Ω
=> current = voltage/ resistance
=> I = 3.3/3 = 1.1 A
hence current will be 1.1 Ampere.
1) resistance of circuit when key K is open
When the key is open, the only resistance left will be the 5 Ω resistance in series with 0.5 Ω source resistance,
hence net resistance = 5 + 0.5 = 5.5 Ω
2) current drawn from cell when key is open
When the key is open, net resistance = 5.5 Ω
=> current = voltage/ resistance
=> I = 3.3/5.5 = 0.6 A
hence current will be 0.6 Ampere.
3) resistance of circuit when key k is closed
When key is closed, the 5 Ω will be in parallel with series combination of 2 and 3Ω
hence net resistance = 0.5 + 5||(2+3)
= 0.5 + 5||5
= 0.5 + 2.5
= 3 Ω
4) current drawn from cell when key is closed
When the key is closed, net resistance = 3 Ω
=> current = voltage/ resistance
=> I = 3.3/3 = 1.1 A
hence current will be 1.1 Ampere.
Hope that is cool