51) This question asks us to find how the brightness of the bulb at R[1] would be affected when the switch is closed. This is the same as finding the power dissipated at R[1].
There are 3 forms of the equation for power. We can use this to our advantage. We can pick the form of the equation with the variables that are easiest to find.
Let’s start the situation where the switch is open. We already have the resistance of R[1] and can easily find the current flowing through R[1], so it will be simplest to calculate the power with the red equation.
In problem 49, we came up with this simplified version of the circuit. We also found out that the current through each resistor in the circuit would be 1.0 Amperes. You can take a look here for background information about how we came up with this. We have all the information necessary to solve for the current.
The second step is to find the power dissipated through R[1] when the switch is closed. We already know R[1], and it will be easier to find the voltage than to find the current, so we will use the purple equation for this situation. This is how the circuit would look when the switch is closed.
Since this is a parallel circuit, the voltage through the top parallel and the bottom parallel will be the same as the total voltage. In the second parallel, the voltage is split evenly amongst R[1] and R[2] because they have the same resistance.
When we plug these values in we will find our power dissipated through R[1].
The power dissipated in both situations is 2 Watts. This means that the brightness in both situations will be the same.
Answer: E
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