mission ec4: resistance, voltage and current\na certain electrical circuit contains a battery, wires and a…

Explanation:

Step1: Recall power - current - voltage - resistance relationships

The power dissipated in a circuit element (like a light - bulb) is given by $P = VI=\frac{V^{2}}{R}=I^{2}R$. A less - bright bulb means less power is dissipated in it.

Step2: Analyze option a

If we increase the voltage $V$ of the battery, from $P=\frac{V^{2}}{R}$ (assuming $R$ is constant), the power $P$ will increase, and the bulb will shine brighter. So option a is incorrect.

Step3: Analyze option b

If we decrease the voltage $V$ of the battery, from $P = \frac{V^{2}}{R}$ (assuming $R$ is constant), the power $P$ will decrease, and the bulb will shine less brightly. So option b is correct.

Step4: Analyze option c

If we increase the resistance $R$ of the circuit, from $P=\frac{V^{2}}{R}$ (assuming $V$ is constant), the power $P$ will decrease, and the bulb will shine less brightly. So option c is correct.

Step5: Analyze option d

If we decrease the resistance $R$ of the circuit, from $P=\frac{V^{2}}{R}$ (assuming $V$ is constant), the power $P$ will increase, and the bulb will shine brighter. So option d is incorrect.

Step6: Analyze option e

If we open the circuit, there will be no current flowing through the bulb ($I = 0$). From $P=VI$, when $I = 0$, $P=0$, and the bulb will not shine at all (which is the ultimate case of shining less brightly compared to when it is on). So option e is correct.

Answer:

B. decrease the voltage of the battery, C. increase the resistance of the circuit, E. open the circuit