
When a switch is in $OFF$ position,
$\left( i \right)$ Circuit starting from the positive terminal of the cell stops at the switch.
$\left( {ii} \right)$ Circuit is open.
$\left( {iii} \right)$ No current flows through it.
$\left( {iv} \right)$ Current flows after some time.
Choose the combination of correct answers from the following.
A. all are correct
B. $\left( {ii} \right)$ and$\left( {iii} \right)$ are correct
C. only $\left( {iv} \right)$ is correct
D. only $\left( i \right)$ and $\left( {ii} \right)$ are correct
Answer
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Hint:A switch is a passive component in an electrical network and the switch is a way of controlling the current in the circuit. Some switches' purpose is just open or close whereas other switches can change the path of the current. An electric current will only flow in a circuit if there is no break in the circuit or else the current in the circuit will be zero.
Complete answer:
When the switch is in the $OFF$ position, the circuit is said to be open and when the circuit is open No current flows in the circuit. For example: consider a circuit consisting of a battery, a bulb and a switch as shown in the figure below,
In the first circuit the switch is in the $OFF$ position, the circuit is said to be open and the current will not flow and the bulb will be turned $OFF$. In the second circuit the switch in $ON$ position the circuit is said to be closed and the current will not flow and the bulb will be turned $ON$.
Hence, option B is correct.
Note:The circuit is said to be open until the switch is open. The closed path will be created in the circuit when the switch. The basic idea beyond switching on of light is you are actually switching on all the internal circuits so as to turn on the bulb.
Complete answer:
When the switch is in the $OFF$ position, the circuit is said to be open and when the circuit is open No current flows in the circuit. For example: consider a circuit consisting of a battery, a bulb and a switch as shown in the figure below,
In the first circuit the switch is in the $OFF$ position, the circuit is said to be open and the current will not flow and the bulb will be turned $OFF$. In the second circuit the switch in $ON$ position the circuit is said to be closed and the current will not flow and the bulb will be turned $ON$.
Hence, option B is correct.
Note:The circuit is said to be open until the switch is open. The closed path will be created in the circuit when the switch. The basic idea beyond switching on of light is you are actually switching on all the internal circuits so as to turn on the bulb.
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