Connecting resistors in series always gives maximum resistance and parallel gives minimum resistance. (a) The highest total resistance is given by 𝑅 =𝑅₁+𝑅₂+𝑅₃+𝑅₄ =4 Ω+ 8 Ω+ 12 Ω+ 24 Ω =48 Ω (b) The lowest total resistance is given by 1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃+1/𝑅₄ 1𝑅=1/4+1/8+1/12+1/24 = 6+3+2+1/24 = 12/2Read more
Connecting resistors in series always gives maximum resistance and parallel gives minimum resistance.
(a) The highest total resistance is given by
𝑅 =𝑅₁+𝑅₂+𝑅₃+𝑅₄
=4 Ω+ 8 Ω+ 12 Ω+ 24 Ω
=48 Ω
(b) The lowest total resistance is given by
1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃+1/𝑅₄
1𝑅=1/4+1/8+1/12+1/24
= 6+3+2+1/24 = 12/24
⟹𝑅=24/12=2 Ω
In parallel there is no division of voltage among the appliances. The potential difference across each appliance is equal to the supplied voltage and the total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel. For more answers visit to website: httpsRead more
In parallel there is no division of voltage among the appliances. The potential difference across each appliance is equal to the supplied voltage and the total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.
Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter of resistance 500 Ω are connected in parallel. The net resistance in parallel is given by 1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃ Here, 𝑅₁=100 Ω, 𝑅₂=50 Ω and 𝑅₃=500 Ω, So, 1/𝑅 = 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.2Read more
Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter of resistance 500 Ω are connected in parallel.
The net resistance in parallel is given by
1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃
Here, 𝑅₁=100 Ω, 𝑅₂=50 Ω and 𝑅₃=500 Ω, So,
1/𝑅 = 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.25 Ω
Now, using Ohm’s law V = IR, we have
𝐼=𝑉/𝑅 = 220 𝑉/31.25 Ω = 7.04 𝐴
Hence, the resistance of electric iron is 31.25 Ω and current through it is 7.04 A.
(a) The net resistance in parallel is given by 1/𝑅=1/𝑅₁ +1/𝑅₂ Here, 𝑅₁=1 Ω and 𝑅₂=10⁶ Ω, So, 1/𝑅 =1/1 + 1/10⁶ = 10⁶+1/10⁶ ⟹ 𝑅=10⁶ /10⁶ +1≈1Ω (b) The net resistance in parallel is given by 1/𝑅=1/𝑅₁ +1/𝑅₂ +1/𝑅₃ Here, 𝑅₁ =1 Ω, 𝑅₂=10⁶ Ω and 𝑅3=10⁶ Ω, So, 1/𝑅 = 1/1+1/10³+1/10⁶ = 10⁶ +10³+1/10⁶ =1001001/1Read more
(a) The net resistance in parallel is given by
1/𝑅=1/𝑅₁ +1/𝑅₂
Here, 𝑅₁=1 Ω and 𝑅₂=10⁶ Ω, So,
1/𝑅 =1/1 + 1/10⁶ = 10⁶+1/10⁶ ⟹ 𝑅=10⁶ /10⁶ +1≈1Ω
(b) The net resistance in parallel is given by
1/𝑅=1/𝑅₁ +1/𝑅₂ +1/𝑅₃
Here, 𝑅₁ =1 Ω, 𝑅₂=10⁶ Ω and 𝑅3=10⁶ Ω, So,
1/𝑅 = 1/1+1/10³+1/10⁶ = 10⁶ +10³+1/10⁶ =1001001/1000000 ⟹𝑅=1000000/1001001= 0.999Ω ≈ 1Ω
The resistivity of an alloy is higher than the pure metal and it does not corrode easily. Moreover, even at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal. For more aRead more
The resistivity of an alloy is higher than the pure metal and it does not corrode easily. Moreover, even at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 ohm, 8 ohm, 12 ohm, 24 ohm?
Connecting resistors in series always gives maximum resistance and parallel gives minimum resistance. (a) The highest total resistance is given by 𝑅 =𝑅₁+𝑅₂+𝑅₃+𝑅₄ =4 Ω+ 8 Ω+ 12 Ω+ 24 Ω =48 Ω (b) The lowest total resistance is given by 1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃+1/𝑅₄ 1𝑅=1/4+1/8+1/12+1/24 = 6+3+2+1/24 = 12/2Read more
Connecting resistors in series always gives maximum resistance and parallel gives minimum resistance.
See less(a) The highest total resistance is given by
𝑅 =𝑅₁+𝑅₂+𝑅₃+𝑅₄
=4 Ω+ 8 Ω+ 12 Ω+ 24 Ω
=48 Ω
(b) The lowest total resistance is given by
1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃+1/𝑅₄
1𝑅=1/4+1/8+1/12+1/24
= 6+3+2+1/24 = 12/24
⟹𝑅=24/12=2 Ω
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
In parallel there is no division of voltage among the appliances. The potential difference across each appliance is equal to the supplied voltage and the total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel. For more answers visit to website: httpsRead more
In parallel there is no division of voltage among the appliances. The potential difference across each appliance is equal to the supplied voltage and the total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
An electric lamp of 100 ohm, a toaster of resistance 50 ohm, and a water filter of resistance 500 ohm are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter of resistance 500 Ω are connected in parallel. The net resistance in parallel is given by 1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃ Here, 𝑅₁=100 Ω, 𝑅₂=50 Ω and 𝑅₃=500 Ω, So, 1/𝑅 = 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.2Read more
Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter of resistance 500 Ω are connected in parallel.
The net resistance in parallel is given by
1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃
Here, 𝑅₁=100 Ω, 𝑅₂=50 Ω and 𝑅₃=500 Ω, So,
1/𝑅 = 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.25 Ω
Now, using Ohm’s law V = IR, we have
𝐼=𝑉/𝑅 = 220 𝑉/31.25 Ω = 7.04 𝐴
Hence, the resistance of electric iron is 31.25 Ω and current through it is 7.04 A.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
Judge the equivalent resistance when the following are connected in parallel
(a) The net resistance in parallel is given by 1/𝑅=1/𝑅₁ +1/𝑅₂ Here, 𝑅₁=1 Ω and 𝑅₂=10⁶ Ω, So, 1/𝑅 =1/1 + 1/10⁶ = 10⁶+1/10⁶ ⟹ 𝑅=10⁶ /10⁶ +1≈1Ω (b) The net resistance in parallel is given by 1/𝑅=1/𝑅₁ +1/𝑅₂ +1/𝑅₃ Here, 𝑅₁ =1 Ω, 𝑅₂=10⁶ Ω and 𝑅3=10⁶ Ω, So, 1/𝑅 = 1/1+1/10³+1/10⁶ = 10⁶ +10³+1/10⁶ =1001001/1Read more
(a) The net resistance in parallel is given by
1/𝑅=1/𝑅₁ +1/𝑅₂
Here, 𝑅₁=1 Ω and 𝑅₂=10⁶ Ω, So,
1/𝑅 =1/1 + 1/10⁶ = 10⁶+1/10⁶ ⟹ 𝑅=10⁶ /10⁶ +1≈1Ω
(b) The net resistance in parallel is given by
1/𝑅=1/𝑅₁ +1/𝑅₂ +1/𝑅₃
Here, 𝑅₁ =1 Ω, 𝑅₂=10⁶ Ω and 𝑅3=10⁶ Ω, So,
1/𝑅 = 1/1+1/10³+1/10⁶ = 10⁶ +10³+1/10⁶ =1001001/1000000 ⟹𝑅=1000000/1001001= 0.999Ω ≈ 1Ω
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
The resistivity of an alloy is higher than the pure metal and it does not corrode easily. Moreover, even at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal. For more aRead more
The resistivity of an alloy is higher than the pure metal and it does not corrode easily. Moreover, even at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/